◉This book is divided into nine chapters. The first chapter [describes] the basic purport of the book; the second [explains] that error occurs in sight because of reflection; the third [focuses] on error that arises in plane mirrors; the fourth [focuses] on error that originates in convex spherical mirrors; the fifth [focuses] on convex cylindrical mirrors; the sixth [focuses] on
convex conical [mirrors]; the seventh [focuses] on concave spherical [mirrors]; the eighth [focuses] on concave cylindrical [mirrors]; the ninth [focuses] on concave conical [mirrors].

1.1◉ It was shown in the preceding books how forms are apprehended in mirrors by the visual faculty, how the lines of reflection or incidence are disposed, [and] how images are disposed and where they are located. However, the form is not always perceived as it actually exists by means of reflection. For in concave mirrors the image of [one’s] face appears distorted, and its proper
disposition is obscured from sight, so it is obvious that error occurs in the perception of forms through reflection. In the present book it is [our] purpose to explain how this error occurs and the reason for it, as well as to discuss the different types of errors due to the different types of mirrors.

2.1◉ The second book showed how forms are perceived in direct vision, and the third book carefully analyzed the particular factors that lead to error in that [kind of] vision when the [conditions for proper vision] exceed or fall short of the [appropriate] threshold. The perception of forms by means of a reflection [of rays] occurs in the same way [as it does] in direct vision, and [so]
the things that are apprehended in direct vision are also apprehended in reflected vision—such things as light, color, shape, size, distance, and the like [i.e., the full range of visible intentions].

2.2◉ Moreover, just as happens in the direct visual apprehension of things [whose forms] are already ensconced [in the soul] and known, so in reflected vision there is a correlation [of the form] to something else [like it] so that a conclusion is drawn and a judgment is made in the soul. Hence, any excesses or defects in the threshold conditions [for proper sight that] cause an error in
direct vision likewise cause [an error] in reflected vision. And according to each case [of excess or defect in the threshold condition], the error is magnified in reflected vision because of the diminished light that results from the weakening caused by the actual reflection.

2.3◉ Furthermore, to generalize, we should say that the proper dispo-sition of the form cannot be perceived in reflected vision as it can be in direct vision because of a threefold constraint specific to reflection. The first is that in reflection the form of the object appears to the viewer to lie directly in front of the eyes when this is not actually the case. The second [is] that the
light and color in the visible object are mingled with the color of the mirror, and the visual faculty perceives that mingled [color] rather than the actual color or light belonging to the visible object. The third is that, as has been pointed out earlier [in book 4], reflection itself weakens light and color, so the actual light and color will be less clearly seen in reflected vision than in direct vision.

2.4◉ In addition, earlier discussions showed that the range of the [limits of the] threshold conditions [whose excess or defect] leads to error depends on the intensity of the light and color, for that range will be greater in stronger light or color [and] less in weaker [light or color]. And, since light and color will be weakened by reflection, the range of the [limits of the] threshold
conditions [whose excess or defect] leads to particular kinds of error will be less in reflected vision than in direct vision, and the shortening of that range leads to an increase in the number of errors. Besides, certain tiny features of objects can be perceived through direct vision that are in no way perceptible through reflected vision. It is therefore evident that reflected vision exceeds direct vision in the degree and number of errors.

3.1◉ In each kind of mirror a misperception of forms occurs, but the variety of errors [that occur] depends on the variety of mirrors [in which the forms are perceived]. In plane mirrors less error occurs than in the others. For in these [kinds of mirrors] the proper shape, spatial disposition and size [of the object] are perceived, just as [they are] in direct vision, which will be shown
by [the following] demonstration.

3.2◉ [PROPOSITION 1] Imagine a plane mirror, and let line AB [in figure 6.3.1, p. 97] on that mirror’s surface be the common section of the mirror’s surface and a plane perpendicular to the mirror’s surface. Let H and Z be two points in that perpendicular plane, [let] E [be] the center of sight, and draw perpendicular HL from point H to the mirror’s surface. Extend it so that LG = LH.
Likewise, extend perpendicular ZF so that DF = FZ.

3.3◉ It is clear from earlier discussions [i.e., book 5, proposition 1, in Smith, Alhacen on the Principles, 399] that [the form of point] H is reflected to [point] E from a point on the mirror, and its image-location G lies as far from the mirror’s surface [below it] as H [lies above it]. By the same token, [the form of point] Z is reflected to [point] E, and its image-location is D.

3.4◉ Now when line ZH is drawn, and likewise line GD, [the form of] any point on line ZH is reflected to [point] E. Its image-location lies the same distance from the mirror’s surface as the point itself, and so any point on line ZH appears to lie the same distance [from the mirror’s surface] as it will [actually] lie [from that surface]. Hence, if line ZH is straight, line DG will be
straight. If it is curved, DG will be an arc of the same curvature, so line ZH will appear the same size and shape as it is, which is what was set out [to be proven].

3.5◉ However, if there are various colors that are only slightly different from one another at points along line ZH, the variation [among them] may not be perceived; instead, a single blend of color will be presented to sight. Hence, because of reflection there will be an error involving light and color, and in addition [an error] concerning number. For that difference among the colors and
lights might be perceptible in direct vision, but the [perceptibility of the] color has exceeded the threshold with respect to reflected vision, although not with respect to direct vision. Likewise, tiny features that could be discerned in direct vision are either hidden or confused in reflected vision.

3.7◉ In the case of spatial disposition an error clearly arises from reflection alone, for in the image we perceive things on the left-hand side of the visible object that we would see on the right-hand side if the object were [actually placed in front of us] at the image-location. For, when something faces something else, its corresponding spatial disposition is opposite because what is
the right-hand side of the one will be the left-hand side of the other. Accordingly, the right-hand side of the visible object is the left-hand side of the image, whereas the left-hand side of the image will be its right-hand side to the viewer, but it is perceived on the left-hand side of the image.⁑

3.8◉ Overall, in the case of light, color, or spatial disposition, error invariably arises from the very reflection itself. In these cases, as well as in others, the things that lead to error in direct vision likewise lead to error in reflected vision, and more easily because the [range of] threshold conditions for each is smaller in reflected vision than in direct vision. One example for
all of these [cases] may be applied, and the same should be understood [to apply] to the rest.

3.10◉ In mirrors, the object will appear smaller than it should at a given distance, whereas at such a distance it may look smaller than it should in direct vision, but not to such a great extent.⁑ And this increased diminution [which happens] in mirrors is due to the decrease in the [range of] threshold conditions [for the perception] of distance.

3.11◉ In [the perception of] shape error sometimes arises in mirrors for the same reasons it does in direct vision, but [it does so] more significantly and more frequently according to spatial disposition.

3.12◉ If a rope or something like it faces a mirror at a given distance, and if its ends cannot be perceived by the visual faculty, it may appear to lie on the very surface of the mirror. The same thing happens in direct vision. If some rope is placed facing a window and the ends of the rope cannot be seen, the separation between rope and window will not be apparent, even if it is
significant, and [this] is due to spatial disposition.⁑ Moreover, if one of the ends is visible but the other is not, that end may appear to lie in the plane [of the window]. In each case, where [error] occurs in direct vision, it occurs likewise in reflected
vision.

4.1◉ The entire range of errors that occur in plane mirrors also occurs in convex spherical [mirrors], and besides this, a visible object looks smaller than it should in [convex] spherical mirrors. Overall, in these [kinds of] mirrors nothing about the visible object is perceived as it actually is except the arrangement of its parts, which appears in the mirror as it actually exists in the
visible object.

4.3◉ Let AB [in figure 6.4.2, p. 97] represent a visible line [on some object, let] ZP be the mirror, D the center of the [great] circle [produced by the plane of reflection on the mirror], and E the center of sight. Let [the form of point] A be reflected to [point] E from point H, and [let the form of point] B [be reflected to E] from point N. When it is extended, line AB will pass
through the center of the mirror, or [it will] not.

4.4◉ Let it pass through. From point N draw line NL tangent to the circle, and from point H [draw] tangent HM. Draw the line[-couple]s of reflection BN and EN, and AH and EH, extend lines EH and EN until they fall on normal AD, and let T and Q be the points where they fall. It is evident that T is the image-location for A, and Q is the image-location for B. I say that AB > QT.

4.7◉ But if line AB, when it is extended, does not reach the center [of the circle], then from point A [in figures 6.4.2a and 6.4.2b, p. 98] draw line AG to the center, let G be the center, and from point B draw line BG. Let point D be the image-location for [point] A, let [point] E be the image-location for [point] B, and draw line ED, which is the image of line AB. I say that [object] AB
> [image] ED because ED is either parallel to AB or not.

4.8◉ If it is parallel [as in figure 6.4.2a], it is clear that it is smaller [i.e., ED < AB because triangles EDG and BAG are similar, so BA:ED = BG:EG, and BG > EG]. If it is not parallel [as in figure 6.4.2b], extend [ED] until it meets AB. Let Z be the [point of] intersection, and from point E draw EH parallel to AB. Angle EDH is acute, right, or greater [than a right angle].

4.10◉ If it is acute, it could happen that the form [i.e., ED] is larger than the object [AB] whose form it is, which, although it may be larger, will happen rarely. And when it does happen, the form may be perceived from such a distance that it will appear smaller than it should because the object itself may appear smaller [than it should] at that distance [in direct vision].⁑

4.11◉ [PROPOSITION 3] It will now be demonstrated that in these [kinds of] mirrors a form may sometimes appear larger than the visible object, i.e., when it [actually] is larger, and that it may be perceived [as larger] from such a distance that its size can be discerned with proper certitude.

4.12◉ Let A [in figure 6.4.3, p. 99] be the center of the mirror, and take a plane of reflection that will cut the mirror along a [great] circle. Let that circle be EDB, let ED be the diameter of that circle, and extend diameter ED to Z so that rectangle EZ,ZD is not greater than AD2, which is clear[ly possible], since it is possible for a line to be added to diameter ED such that the
rectangle formed by the whole and the added part equals AD2 [by Euclid, III.36].⁑ Bisect line ZD at point H. Hence, AH will be half of EZ. Accordingly, [since AD < AH, which = half EZ, while DH =
half DZ] rectangle AD,HD will not be greater than one-fourth AD2 [because it is no greater than one-fourth rectangle EZ,ZD, which is no greater than AD2], and since AH,HD > HD2, let AH,HT = HD2.⁑

4.13◉ Produce a circle according to length AH [as radius], and from point H draw chord HQ equal to one-half line HD. Draw lines QA and QT, and at point Q form angle HQN equal to angle QAH. Accordingly, since these two angles in these two triangles [HQN and QAH] are equal, and since one [angle], i.e., QHA, is common, the third [angle] = the third [angle], i.e., [angle] AQH = angle HNQ. And
[so] the triangles will be similar [by Euclid, VI.4], and [according to proportional sides] AH:HQ = HQ:HN. Therefore, AH,HN = HQ2 [by Euclid, VI.17].

4.15◉ Angle QHD is acute, however, and [it is] equal to angle HQA, since they are subtended by equal sides in the larger triangle [i.e., QA and HA, which are radii of the larger circle]. Therefore, angle QHN [in triangle AQH] = angle HNQ [in triangle HQN, which is similar to triangle AQH, by previous conclusions], and so HQ = QN.

4.16◉ Also, angle HNQ is acute, so [adjacent] angle QNT is obtuse. Hence, TQ2 exceeds QN2 + TN2 by TN,NH because, as Euclid claims [in II.12], the square on the opposite side of an obtuse [angle] exceeds the squares on the [other] two sides by twice the rectangle formed by one of the sides and the adjoining segment that extends to where the perpendicular is dropped [to it] from the
endpoint of the other side. And if a perpendicular is dropped from point Q to line HT, it will fall at the midpoint of line HN [because triangle QNT is isosceles], and twice the rectangle formed by TN and one-half HN equals TN,HN.

4.20◉ Hence, IA will be the mean proportional between CA and HA, whose converse we touched upon a bit earlier.⁑ Accordingly, CA:IA = IA:HA, and the remainder will be in the same proportion to the remainder, i.e., CI:IH [= CA:IA = IA:HA], and since IA > HA, CI > IH.

4.23◉ But HT < one twenty-fifth AH [by previous conclusions], and its half < half of one twenty-fifth. But line AH is divided into twenty-five parts, [so] two-thirds [of those 25 parts, i.e., < 17] + half of one twenty-fifth [part] does not add up to 18 parts. Therefore, IH:HT > 25:18.

4.27◉ From the fifth [proposition] of the fifth book you will [then] have that angle IBZ2 = angle HBA, for, since IL:HL = IA:AH [by construction], H will be the image-location [of object-point I] in the case of reflection from point B [when HB, extended beyond B to H’, forms the line of reflection].⁑ And if the contrary is claimed, so that some other image-location is chosen, you will disprove it by a reductio ad absurdum, given that it is impossible for the ratio of IA to the line from point A to the image-location not to be as [the ratio of] IL to the line from point L to the image-location.

4.28◉ Therefore, since H is the image-location, and since LB is tangent [to the mirror] on AB, then when it is extended [to H’], HB will form an angle of reflection [H’BZ2] equal to its vertical [angle HBA], and because LB is perpendicular to ABZ2, it will follow that angle IBL = angle LBH. By the same token, angle IGZ1 = angle TGA, and since MG is perpendicular [to AG], angle IGM = angle
MGT.

4.29◉ Now draw line HP from point H to line AB parallel to [line] IB, and from point T [draw] TR parallel to IG. Angle IBZ2 = angle HPB. But, as was claimed [earlier], angle IBZ2 = angle HBA, and so the two angles HBA and HPB are equal, so the two sides HB and HP [of triangle HBP] are equal. Likewise [in triangle TGR, side] TR = [side] TG. But angle HPB is acute, since it is equal to the
angle of reflection [IBZ2, so adjacent] angle HPA will be obtuse, and HA > HP [by Euclid, I.19], so it will be greater than HB [since HB = HP]. So too, TA > TG.

4.31◉ But since angle HPA [in triangle HPA] is obtuse, HA2 will exceed (HP2 + AP2) by twice the rectangle formed by AP and the line [segment] extended from point P to the perpendicular dropped [to AP] from point H [by Euclid, II.12]. But the perpendicular dropped from point H [to AP] will fall to the midpoint of line PB, since HB and HP are equal [as established earlier], and so HA2 will
exceed (HP2 + AP2) by AP,PB [i.e., HA2 – HP2 – AP2 = AP,PB, so HA2 – HP2 = AP2 + AP,PB]. Accordingly, AH2 exceeds HP2 by AB,AP [i.e., AH2 – HP2 = AB,AP] because (AP,PB + AP2) = AB,AP. Likewise, AT2 exceeds TR2 by AG,AR, or AB,AR, which is identical [since AG = AB, both being radii of the circle].

4.34◉ Now bisect arc BG at point O [see inset to figure 6.4.3a, p. 100], drop the three perpendiculars BF, OY, and GK to line HA, draw line GS from point G parallel to HA, and drop perpendicular BX from point B to AG. If BX were extended to the circle [DGB], line AG would bisect it, as well as the arc whose chord it would form [when extended]. Accordingly, it[s other half after extension]
would cut off another arc equal to arc BG, since its other arc would subtend angle GBX, and so angle GBX is half the angle [GAB] at the center [of the mirror] subtended by that same arc, according to Euclid [III.20]. Hence, angle GBX is one-half angle BAG, which line OA bisects. Therefore, angle GBX = angle OAG. Moreover, the two angles BSG and BXG are right [by construction].

4.35◉ If a circle is imagined on BG [as diameter, and if it] passes through S, it will pass through X [by Euclid, III.31], and arc SX will be formed such that the two angles XBS and XGS [i.e., AGS, since X lies on AG] will fall upon it. These two angles will therefore be equal [by Euclid, III.27]. But angle GAY = [alternate] angle XGS because of the parallelism of lines [AY and GS], so
angle GAY = angle XBS. And, as has [already] been claimed, angle GBX = angle OAG. Angle OAY = angle GBS, and [so] triangle OAY will be similar to triangle GBS.⁑ Hence, GB:BS = OA:AY.

4.44◉ Also, the two triangles AGM and MGK have a common angle [i.e., AMG], and both of them have a right angle [i.e., MGA and MKG]. Hence, they are similar [by Euclid, VI.4], so MK:KG = MG:GA, and so MK < KG [since MG < GA, by previous conclusions]. And since OY > GK, HD < OY [since HD < GK < OY].

4.48◉ Now because of the lack of letters for designating the key points, let us revise the diagram to avoid the excessive tangle of lines. Accordingly, since IA = the line we have designated as AS, construct circle [NRZ in figure 6.4.3b, p.101] according to their length [as radius]. Let us replace S with the letter N, let AG and AB be extended to [points R and C on the circumference of]
this circle, let [the resulting lines] be ABC and AGR, and let us replace the letter Q’ with F. It has been claimed [earlier] that [arc] DF [formerly DQ’] > arc BG. Let arc BM = arc DF, and draw line AMU, as well as lines IM, NM, and [draw] line QM, and extend it to the outer circle. Let it fall to point Z, and draw lines ZA and ZG.

4.49◉ Since arc BM = arc DF, then if common arc [DM] is added [to both], arc MF = arc DB. [Accordingly] angle NAM = angle IAB, the [corres-ponding] sides [NA and IA, and MA and BA, of triangles NAM and IAB] will be equal [and will contain equal angles, so the triangles will be equal, by Euclid, I.4], and [so] MN = IB. And since AQ was assumed earlier to be equal to AH [because they are
radii of the same circle passing through Q and H, sides] AQ and AM [of triangle AQM] = [corresponding sides] HA and AB [of triangle HAB], and angle [QAM contained by sides AQ and AM in triangle AQM is equal] to angle [HAB contained by equal corresponding sides HA and AB in triangle HAB, so the two triangles are equal, by Euclid, I.4]. [Hence] QM = HB, and angle QMN = angle HBI, since both the sides containing them [i.e., QM and NM, and HB and IB, respectively] are
equal [as concluded earlier]. [Accordingly] base IH = base NQ, and angle NMU = angle IBC.

4.50◉ But angle IBC [which is IBZ2 in figure 6.4.3a] = angle HBA [by previous conclusions], and angle HBA = angle QMA, [so] angle NMU = angle QMA. And since QMZ is a straight line, as we stipulated, angle QMA = [vertical] angle UMZ, so [angle of incidence NMU = angle of reflection UMZ, and so the form of] point N is reflected to [point] Z from point M, and its image-location is [point] Q.
This still falls short of a proof to show that all of [line of reflection] MZ lies outside the circle, which will be demonstrated as follows.

4.51◉ It is clear that the tangent drawn from point B will fall between I and H [since HBA and IBC are acute, by previous conclusions], and [it is clear] that point B lies as far from point H as point M lies from point Q, and IH = NQ [from previous conclusions]. Therefore, the tangent drawn from point M will fall between N and Q. QM therefore intersects the circle, so all of MZ [lies]
outside the circle, and so what we set out [to be proven is demonstrated].⁑

4.53◉ Angle BAG will be equal to, less than, or greater than angle GAM.⁑ Let it be equal [as represented in figure 6.4.3c, p. 102]. Accordingly, if angle [C]BAG[R] is subtracted from angle IAB[C], and angle [U]MAG[R] from angle ZAU, angle IAG[R] will be left equal to angle ZAG[R]. [Thus], IG = ZG, triangle [IAG equals] triangle [ZAG], and angle IGA = angle ZGA. It will follow that angle IGR = angle ZGR. But angle IGR = angle TGA [from previous conclusions]. [So] angle TGA = angle ZGR. Hence, if TG is extended, it will
reach [point] Z [on the circle passing through points I and N in figure 6.4.3c], so TGZ is a straight line. [The form of point] I is therefore reflected to [point] Z from point G, and [point] T is its image-location.

4.54◉ So let Z be the center of sight. [The forms of] the two points N and I will be reflected to it from the two points M and G, and the [respective] image-locations will be points T and Q. Thus, TQ will be the image of line IN, and it was proven earlier that TQ = IN [i.e., IS in figure 6.4.3a], and so it can happen in these kinds of mirrors that the image is the same size as the visible
object.⁑

4.55◉ If, however, angle BAG > angle GAM [in figure 6.4.3d, p. 103], then angle ZAG > angle IAG. Let angle KAG = angle IAG. Since point K is lower than [i.e., to the left of] point Z, and point M is lower than [i.e., to the left of point] G, line KG will intersect line ZM. Let it intersect at point L. Accordingly, if the center of sight lies at point L, [the form of point] N is
reflected to it from point M, and [point] Q is its image location; [the form of point] I is reflected to it from point G, and [point] T is its image-location, according to the preceding proof. And so TQ is the image of IN, which is what was set out [to be proven].⁑

4.56◉ On the other hand, if angle BAG < angle GAM [as in figure 6.4.3e, p. 104], angle ZAG < angle IAG. Let angle OAG = angle IAG, and extend line OG. It is clear that [the form of point] I is reflected to [point] O from point G. Line OG will either intersect line ZMQ outside the [great] circle [FDGB] of the mirror, or it will not.⁑

4.58◉ If line OG should happen to intersect line ZMQ inside the circle [as represented in figure 6.4.3e, p. 104, where X is the intersection-point], the foregoing proof cannot be applied. Instead, I say that a point can be found outside that entire surface to which [the forms of] the two points I and N are reflected from two points [on the mirror], and [that] TQ [forms] the image.

4.59◉ For instance, from the foregoing it is evident that angle NAZ is twice angle IAB [since angle IAB = angle NAM = angle MAZ, by construc-tion], and angle IAO is twice angle IAG [since angle IAG = angle OAG, by construction]. Furthermore, angle NAZ exceeds angle IAO by an amount no greater than angle NAI.⁑ In addition, the two angles OAI and ZAN [together] are greater than the third [angle], which is IAN, the two [angles] OAI and IAN are greater than the third [angle] NAZ, and the two [angles] ZAN and NAI are greater than the third [angle] IAO. We therefore have three angles [IAO, IAN, and NAZ], any two of which [together] are greater than the third.

4.60◉ From these [angles], therefore, a solid angle can be formed [by Euclid, XI.23].⁑ Form that angle at A [in figure 6.4.3f, p. 105], let line SA be erected at A, let angle IAS = angle IAO, [and let] angle NAS = angle NAZ. Angle NAI will remain where it is, and line AS will be formed equal to lines AN and AI, which are both equal.⁑

4.62◉ Bisect angle TAS with line AY, and let Y be the point at which that line will intersect line TS. Since angle IAG is one-half angle IAO, it is evident that angle TAG = angle TAY, whereas angle GTA = angle YTA, and one side, i.e., TA, is common [to both triangles TAG and TAY. Accordingly] TG = TY, triangle [TAG] = triangle [TAY], AY = AG, and so Y will lie on the surface of the sphere
[from which the mirror is formed]. Thus, angle IAG = angle IAY, and the [corresponding] sides [IA and AG] = the [corresponding] sides [IA and AY. So] triangle [IAG] = triangle [IAY], [angle] AGI = angle AYI [and] line IY in its full extent = [line] IG.⁑

4.64◉ But line AS intersects the sphere [of the mirror]. Let O’ [in figure 6.4.3h] be the point of intersection. Accordingly, the three points O’, Y, and D lie on the surface of the sphere, so line O’YD is a segment of a [great] circle of the sphere, and it is the common section of the sphere’s surface and plane ITASP, so [the form of] point I is reflected to point S from point Y [within
plane of reflection ITASP], and T is the image-location.

4.65◉ Likewise, if angle NAS [in figure 6.4.3f. p. 105] is bisected by [line] AZ’Z«, it will be proven in the preceding way that QZ’ = QM, AZ’ = AM, Z’S = MZ, and the two angles NZ’Z« and SZ’Z« are equal to the two angles NMU and ZMU. And so [the form of point] N is reflected to [point] S from point Z’, and Q is the image-location, so TQ is the image of IN, which is what was set out [to
be proven].

4.66◉ Now if a perpendicular is dropped from point I to NA, it should fall between N and Q, not beyond N because angle INA is acute, since it is equal to angle NIA, and if that perpendicular were to fall beyond N, an acute [angle] would be greater than a right angle.⁑ Therefore, that perpendicular will form a right angle on NQ, and that angle will be subtended by line IN, so line IN > that perpendicular, and so that perpendicular < TQ [since TQ is equal to IN by the initial construction].

4.67◉ The [form of the] point on line NQ where the perpendicular falls is reflected to point S, and its image will lie on line NA above point Q because the farther away [from the mirror’s surface] the points that are reflected lie, the more their image-locations approach the center of the circle, according to the tenth [proposition] of the fifth book.⁑

4.68◉ Moreover, any line drawn from point T to any point on NQ above Q will be longer than TQ. Therefore, the image of the perpendicular will be longer than the perpendicular itself [which is shorter than IN]. By the same token, no matter what line is drawn to NQ from point I between this perpendicular and IN, its image will be longer than it.

4.69◉ But these claims may be determined more definitively [as follows. The form of] point N [in figure 6.4.3m, p. 108] is reflected to [point] Z from point M, and Q is the image-location. Line QM cuts the circle at point E. Therefore, the tangent drawn from point Z to the circle will fall at some point on arc ME, and that tangent will fall above Q, since the point where it will fall [on
NQ] will form the endpoint of tangency [X on cathetus NA] and [thus] the limit of images, and points below that endpoint of tangency cannot be reflected, [whereas points] above it can.⁑

4.70◉ Therefore, if the perpendicular dropped from point I falls above the endpoint [X] of tangency, the point where it falls is reflected, and the image of the perpendicular will be longer than the perpendicular [itself]. But if the perpendicular should fall at or below the endpoint of tangency, the point where it falls is not reflected, so there will be no image of the perpendicular.
However, since the endpoint of tangency lies below N, there will be an infinitude of points between the endpoint of tangency and N, and any of them will be reflected, and the image of any of them [will lie] on NQ. And the image of any line drawn from point I to any of those points will be longer than the line of which it is the image.

4.71◉ In these [sorts of] mirrors, then, the image may sometimes be the same size as the visible object and sometimes larger, which is what was set out to be explained. Moreover, we have not read an explanation of this matter in any text, nor have we heard anyone who has discussed it or thought about it.

4.72◉ Moreover, in these [sorts of] mirrors straight lines appear curved, such that, in many cases, the curvature [of their images] does not correspond to that of the mirror but is opposite. Likewise, curved [things] will appear curved in these [sorts of] mirrors, and if the curvature corresponds to that of the mirror, it will appear in an opposite orientation, but this must be understood
not [to hold] in all cases but in several, [but] for the sake of explaining this, certain preliminary points must be set out, one of them being as follows.

4.73◉ [PROPOSITION 4, LEMMA 1] If two points lie the same distance from the center of the mirror and different distances from the center of sight, the image of the point lying farther from the center of sight will lie farther from the center of the sphere [forming the mirror] than the [image of the point] lying nearer [the center of sight], and the endpoint of tangency for the farther
[point will lie] farther from the center [of the circle] than the endpoint of tangency for the nearer [point, and this will be the case] whether those points lie in the same plane as the center of sight or in different planes.

4.74◉ The proof [is as follows]. Let T and D [in figure 6.4.4, p. 109] be two points equidistant from G, the center of the mirror, [and let] E be the center of sight. Plane DGT will cut the mirror along [great] circle AB. Let angle EGD = angle TGZ, [let] angle EGT = angle TGH, and find point Q on the circle from which [the form of point] T is reflected to [point] Z [by book 5, proposition
25, in Smith, Alhacen on the Principles, 427-432].⁑ I say that [the form of point] T is not reflected to [point] H from any point on BQ.

4.75◉ It is obvious that [it does] not [do so] from point Q [itself]. Moreover, if some point is taken on BQ, the line [of reflection] drawn to that point from point H will intersect line QZ. [The form of point] T is therefore reflected to that point of intersection from the point selected on BQ, and it is [also] reflected to that same point of intersection from point Q. So [the form of]
point T is reflected to the same point from two points on that circle, which is impossible in these [sorts of] mirrors, as was shown in [proposition 16 of] the fifth book [in Smith, Alhacen on the Principles, 412-414].

4.76◉ It follows that [the form of point] T may be reflected to H from some point on QA. Let that [point] be M [as found by book 5, prop. 25], and from point M draw MN to line GT tangent to that circle. N will be the endpoint of tangency for T with respect to H [as center of sight].⁑

4.77◉ Then from point Q draw tangent QO, which will necessarily lie below MN. Extend ZQ until it falls on GT at point C. C will be the image-location [of T] for Z [as center of sight]. Thus, GT:TO = GC:CO [by book 5, proposition 7, in Smith, Alhacen on the Principles, 404]. So GT:TN > GT:TO. A fortiori, then, GT:TN > GC:CN. Accordingly, let GT:TN = GL:LN. GL > GC, and L will be
the image-location [of T] for [center of sight] H [according to book 5, prop. 7].

4.78◉ So let lines HG, EG, and ZG be equal, [let] GF = GC, [and let] GS = GO. Therefore, since angle EGD = angle TGZ [by construction], and since D lies as far from point E as Z does from point T [given that DG = TG, and EG = ZG, by construction], the image of D with respect to G will lie as high on line GD as the image of T on line GT [with respect to G, by book 5, proposition 17, in
Smith, Alhacen on the Principles, 414-415]. Thus, the image of [point] D [with respect to G lies] at point F [since GF = GC, by construction]. Likewise, the endpoint of tangency for D with respect to E will lie at the same height as the endpoint of tangency [at point O] for Z, so the endpoint of tangency for D [lies] at point S [since GS = GO, by construction].

4.79◉ But since angle EGT = angle TGH [by construction], and since HG = EG [also by construction], L will be the image of T with respect to E, just as it is with respect to H. And N is the endpoint of tangency with respect to E, so the image [L] of the point farther from E [i.e., T] lies farther from the center than the image [F] of the nearer [point D], and the endpoint of tangency [N]
for the farther [point T lies] farther from the center than the endpoint of tangency [S] for the nearer [point D], which was what was set out [to be proven].

4.80◉ [PROPOSITION 5, LEMMA 2] Furthermore, given line AB [in figure 6.4.5, p. 109] divided at points G and D such that AB:BD = AG:GD, I say that, if three lines, i.e., GE, DE, and BE, are drawn from the points of division to intersect at one point, and if a line is drawn from point A to intersect those three lines, that line will be cut according to the aforesaid proportion.

4.82◉ From point H draw HQ parallel to AB. It is clear that AB:BD is compounded from AB:HQ and HQ:BD [i.e., AB:BD = (AB:HQ)(HQ:BD)]. But since QH is parallel to AB, triangle TQH will be similar to triangle BTA, and [so] AB:QH = AT:TH. Likewise, triangle QEH is similar to triangle BED. Therefore, QH:BD = HE:ED. Hence, AB:BD is compounded of AT:TH [which = AB:QH] and HE:ED [which =
QH:BD].

4.83◉ Extend QH until it falls on EG at point M. AG:GD is compounded from AG:HM and HM:GD. But since angle EMH = [alternate] angle ZGD, then angle HMZ [adjacent to EMH] = angle ZGA [adjacent to ZGD], and [so] triangle AZG will be similar to triangle HZM, and AZ:ZH = AG:HM.

4.84◉ But triangle HEM is similar to triangle GED. [Hence] HM:DG = HE:ED. Accordingly, AG:GD is compounded from AZ:ZH and HE:ED, and AG:GD = AB:BD [by construction].⁑ Thus, that same [proportion AG:GD] is compounded from AT:TH and HE:ED, and it is likewise compounded from AZ:ZH and HE:ED. Hence [if we drop the common term HE:ED], AT:TH = AZ:ZH, and so what was set out [has
been proven].

4.85◉ The same proof will hold no matter what line is drawn from point A to intersect those three intersecting lines. And if three other lines are drawn from the three point G, D, and B to intersect at some point other than E, and if any line is drawn from A to intersect those [three lines], it will be cut according to the aforesaid ratio. And so, however the three lines may intersect, if
the [resulting] three lines [represented by] EG, ED, and EB are extended beyond the three points B, D, and G, on the other side [away from the point of intersection], and if lines are drawn from point A to intersect them on that other side, those lines may never be cut according to the aforesaid ratio.⁑

4.86◉ [PROPOSITION 6, LEMMA 3] Moreover, given line AB [in figure 6.4.6, p. 110] cut in the preceding way, if another line, such as AT, is drawn from point A so as to be cut according to the same ratio, and if lines are drawn from the points of division on AB to the points of division on AT, those lines not being parallel, I say that those three [lines] will intersect at the same
point.

4.87◉ The proof [is as follows]. Let AT:TH = AZ:ZH. BT and DH are not parallel, so they will intersect at point E. Line GZ will either intersect [them] at the same point, or it will not. If [it does intersect] at that point, we have what was set out [to be proven]. If not, then draw line EG. It will intersect line AT in some point other than Z. Let that point be L. Thus, AT:TH = AL:LH,
according to the previous proof. But it has been supposed that AT:TH = AZ:ZH, and so it is impossible [for EG to intersect AT at some point other than Z].

4.88◉ Likewise, if it is supposed that line GZ intersects DH at point E, it will be proven in this way that line BT will intersect [it] at the same [point]. So too, if it is supposed that GZ and BT intersect at point E, it will be indisputable that DH will intersect at the same [point].

4.89◉ [PROPOSITION 7, LEMMA 4] Furthermore, given that AB [in figure 6.4.7,p. 110] is divided according to this ratio [AB:BD = AG:GD], if lines GZ, DH, and BT are parallel, and if AT is drawn to cut them, AT will be divided according to this ratio.

4.90◉ The proof [is as follows]. Since DH is parallel to GZ, AZ:ZH = AG:GD, and since BT is parallel to DH, AB:BD = AT:TH. But AB:BD = AG:GD, [so] AT:TH = AZ:ZH, and so [we have demonstrated] what was set out [to be proven]. With these points established, let us proceed to what was proposed [in paragraph 4.72, p. 175-176 above].

4.91◉ [PROPOSITION 8] First of all, it must be shown how in these [sorts of] mirrors the image of an arc is curved with a curvature that accords not with the [surface of the] mirror but with its center.

4.92◉ For instance, let AB [in figure 6.4.8, p. 110] be an arc facing the mirror [composed from sphere YZ on whose surface Y’Z’ is a great circle within the plane of arc AB], let G be the center of that arc as well as of the mirror, [and let] D [be] the center of sight. Draw lines DG, AG, and BG. Take E at random on arc AB, and draw line EG. Let line DG not lie in plane ABG. Line DG will
either be perpendicular to plane ABG, or [it will be] inclined [to it].

4.93◉ Let it be perpendicular. Angles DGA, DGE, and DGB will be equal, and the [corresponding] sides [of triangles DGA, DGE, and DGB will be equal] to the [corresponding] sides, so the bases [DA, DE, and DB will be] equal. Hence, all the points on arc AB will lie the same distance from the center of sight, so the images of all of them [will lie] the same distance from the center.

4.94◉ Let Q, M, and L be the images of A, E, and B. Accordingly, GQ will be equal to GM and GL, so QML will be an arc, and its convex curvature accords with the center [of curvature], not with the [surface of the] mirror, or with the points of reflection, which is what was set out [to be proven].⁑

4.95◉ If, however, line DG is not perpendicular to plane AGB [as in figure 6.4.8a, p. 111], and if a perpendicular [DX] is dropped from point D to that plane, then, since that perpendicular is the shortest of all [possible] lines extending from point D to this plane, the angle this perpendicular forms with respect to G will be smaller than any angle imagined at point G formed by any other
line drawn from point D to the plane, and the farther the line drawn from point D to the plane will lie from the perpendicular, the longer it will be and the greater the angle it will form. Accordingly, if this perpendicular does not fall on arc AEB but on one side of it, all the lines drawn from point D to this arc will be slanted to one side, and the ones that lie farther away will be longer and will form a larger angle.⁑

4.96◉ Let [this] be [the case], then, and take three points, i.e., E, C, and B, on the arc [in figure 6.4.8b, p. 111].⁑ Let L be
the endpoint of tangency for point B, and let M be the endpoint of tangency for point C, for, since C is nearer than B to D, M will be nearer than L to G, and so CM > BL.⁑

4.97◉ Let Q be C’s image, let T be B’s image, and draw TQ. Then draw lines CB and ML, which will intersect when extended, for if a line were drawn from M parallel to CB, it would cut a line from GB equal to CM [and thus be shorter than GL, which means that angle MLG < angle CBG]. Let them intersect at point O.

4.98◉ Since GC:CM = GQ:QM [by book 5, prop. 7], and likewise, since BG:BL = GT:TL [by the same proposition], line QT will intersect lines CB and ML [all at the same point]. Let the intersection be at point O.⁑

4.99◉ Let N [in figure 6.4.8b, p. 111] be the endpoint of tangency for point E. Since point N is lower than point M [by proposition 4, lemma 1], EN > CM. Hence, if they are extended, lines EC and NM will intersect. Let the intersection be at point P, draw line QP, and extend it until it falls upon EG at point F. Then extend line [O]TQ to EG, and let it fall at point K.

4.100◉ It is evident that K will lie above F. However, since GC:CM = GQ:QM, and since three lines [EP, NP, and FP] are drawn through the points of division to meet [at point P] when extended to the other side, they will cut line EG according to the previous ratio [by proposition 6, lemma 3], so GE:EN = GF:FN. But N is the endpoint of tangency [for E, by construction], so F is the
image-location [for E]. Hence, line FQT will be the image of arc ECB, and it will be a curved line rather than a straight one, for TQK is straight [by construction], and the curvature of the [resulting] line [TQF] is not oriented with [that of] the mirror.

4.101◉ Likewise, if the perpendicular dropped from point D falls on the other side of the arc [i.e., to the left of perpendicular D’G], the proof will be identical. But if the perpendicular falls on the midpoint of arc AB, the lines drawn to the arc from point D to opposite sides and equidistant from the perpendicular will be equal and will form equal angles at G. Moreover, their images
will be equidistant from G, and so will the endpoint[s] of tangency,46 and it can be proven in the foregoing way for either side of the arc by itself, according to how it is cut by the perpendicular, that its image is a curved line in the way prescribed, which is what was set out [to be proven].

4.102◉ [PROPOSITION 9] Now take a circle [containing arc ADB in figure 6.4.9, p. 112] whose center [F] is not the center [G] of the mirror [containing arc HK]; nonetheless, let it lie in the same plane as the mirror’s center. I say that, if an arc [ADB] is taken on the [larger] outer circle on the side of the mirror’s center, i.e., nearer that center [G and thus directly opposite it],
its image will be curved.

4.103◉ For, given this arc, draw a line [FG] from the center of the mirror to the center of the outer circle, and extend this line to the given arc [ADB]. The line drawn from the center of the mirror to this arc [i.e., GD], which is a segment of the diameter [FD] of the larger circle, will be shorter than all the [other] lines drawn from the same centerpoint of the mirror to that arc.
Moreover, two equal lines [GA and GB] can be drawn to the given arc from the mirror’s centerpoint on opposite sides of this shortest line, and they will of course be longer [than it]. And if a circle is drawn according [to the length of] either of them from the center of the mirror, arc [AEB on it] will pass through the endpoints of these two lines, and it will be longer [and of a sharper curvature] than the given arc [ADB].

4.104◉ It is clear that the image of this longer arc will be a curved line, according to previous conclusions [in proposition 8]. [It is] also [clear that] the images of the points [A and B] common to this arc [AEB] and the given arc [ADB will be] the same and [that] the midpoint [E] of the longer arc lies farther from centerpoint [G] than the [mid]point [D] on the given arc that
corresponds to it, so its image [will lie] nearer to the centerpoint [G] than the image of point [D] on the given arc that corresponds to it [by book 5, prop. 17, in Smith, Alhacen on the Principles, 414-415]. And so, the image of any point on the outer arc [will lie] nearer the centerpoint than the image of the point on the given arc that corresponds to it. Accordingly, the image of the given [less sharply curved] arc [ADB] is more sharply curved than the image
of the [more sharply curved] outer arc [AEB], so the image of the given arc is curved, which is what was set out [to be proven].

4.106◉ Let A[C]B [in figure 6.4.10, p. 112] be the [straight] line that is seen, [and let] G [be] the center of the mirror. Draw lines AG and BG. They are either equal or not. If [they are] equal, then construct [the] circle [containing arc] AEB on centerpoint G according to their length. Line A[C]B will obviously fall inside [this] circle. From the previous [proposition] it is clear
that the image of arc AEB will be curved. Accordingly, let its image be ZTH. Let Z be the image of A, let H be the image of B, and let T be the image of E.

4.107◉ Extend line GE to cut AB at point C. It is clear that E lies on the same line with C [and] farther from the centerpoint [G] than C. Its image will [therefore] lie nearer to centerpoint [G] than the image of C [by book 5, prop. 17]. So let it be M. It is clear that line ZMH is the image of line AB, and it is a curved line, which is what was set out [to be proven].

4.108◉ [PROPOSITION 11] On the other hand, if lines AG and BG are not equal, then, when it is extended, line AB will either intersect the mirror or not. Let it not intersect [as in figure 6.4.11, p. 113], let AG > BG, construct circle AEQ at [centerpoint] G according to the length of AG [as radius], and extend AB until it touches the circle on the side of B. Let it fall at point
Q.

4.109◉ It is clear from the foregoing [analyses in propositions 8 and 9] that the image of arc AE is curved. Let Z be the image-point for A, and let M be the image-point for E. ZM will [therefore] be the image of arc AE, and since the image of point B lies farther from centerpoint [G] than the image of point E [by book 5, prop. 17], the image [TNZ] of line AB will be curved, [and] this
can be demonstrated according to the midpoints of arc AE and line AB, which is what was set out [to be proven].

4.110◉ Note that in the preceding figure, if a segment is cut from line AB on the side of A, and a segment is cut on the side of B equal to it, the remaining portion of the line will have a curved image, and the proof [of this] will be the same as it is for [the whole of] line AB. Also, if in this figure another segment of line AB is cut on the side of B, the same proof will hold for the
remainder [of the line] as holds for [the whole of] line AB.

4.111◉ [PROPOSITION 12] But if line AB touches the mirror, it will either intersect it or be tangent to it. Let it be tangent [as in figure 6.4.12, p. 114]. Let G be the center of the mirror [containing arc PE in gray], and draw lines AG and BG. Plane ABG cuts the mirror along the circle SEZ [that forms their] common [section]. It is clear that line AB will be tangent to the mirror on
this circle. Let it be tangent at point E. Accordingly, extend AB to E. Let D be the center of sight. The plane containing lines DG and AG cuts the mirror along a [great] circle [that forms the] common [section] of the plane and the mirror. Let ZP be an arc on that circle. Likewise, let HP be an arc on the [great] circle [that forms] a common section of the plane containing DG and BG [and the mirror].⁑

4.112◉ It is clear [from proposition 4, lemma 1, paragraphs 4.74-76] that [the form of point] B is reflected to [point] D from some point [F’] on arc HP. If a tangent is drawn from that point, it will intersect line BG, and the point of intersection will be the endpoint of tangency [for point B on cathetus BG]. Let M be that point [on the resulting tangent F’M].

4.113◉ It is also clear that, if a tangent is drawn from point M to circle SEH, that tangent will fall in front of E because AB is tangent at point E, and point B lies above point M. It will therefore fall at point F, and, when it is extended, the tangent [MF] will intersect line AE. Let it intersect at point T. It will intersect line AG on the other side. Let it intersect at point
C.

4.114◉ Form angle BGS = angle BGD, and extend GS to point L so that it is equal to line DG. Accordingly, arc HS = arc HP, and just as [the form of point] B is reflected to [point] D from a point on arc HP, it[s form] is reflected to L from some point on arc HS. Moreover, the reflection will occur from point F just as the reflection on arc HP occurs from the point [F’] from which the
tangent [F’M’] is drawn to point M, and those two points lie the same distance from point M [which means that the reflections from B to D and from B to L are perfectly equivalent]. So draw lines BF and LF.

4.115◉ [The form of point] A is reflected to D from some point [R] on arc ZP. However, in triangle HZP the two arcs HZ and HP are longer than the third, i.e., ZP.⁑ But HP = HS. Therefore, ZP < ZS. Cut an equal segment from ZS at point Y [so that ZY = ZP], and draw line GY, which, when it is extended to the same length as GD, will necessarily intersect line FL. Let it intersect at point X, and let GXK = GD.

4.116◉ It is clear that, just as [the form of point] A is reflected to D from some point [R] on arc ZP, it is likewise reflected to K from some point [R’] on arc ZY.⁑ I say that it is reflected to it [i.e., K] only from a point that is below F on the side of Z.

4.117◉ For if it is claimed that it can [be reflected] from point F or from some point on arc FY, the line drawn from point A to the point of reflection will intersect line BF. [The form of] point K is reflected to that point of intersection, and [the form of] point L is reflected to the same point, and so two points are reflected in these [sorts of] mirrors to the same point on the same
side, which is impossible [by book 5, prop. 16]. It follows that [the form of] point A is reflected to [point] K from some point [R’] on arc ZF.

4.118◉ If a tangent is drawn from that point, it will intersect line AZ, and it will fall between C and Z because point F is lower than any [other] point on arc ZF, so the tangent from point F is higher than the rest that are drawn from points on arc ZF. So let that tangent [R’N] fall at point N, and draw line NM, and since it passes through the vertex of triangle BMT and cuts the angle
when extended, that line will necessarily intersect BT. Let it intersect at point Q, and draw line GQ.

4.119◉ Now let I be the image of point A; let O be the image of point B; and let U be the image of point Q. Since B lies nearer than A to point G, O will lie farther than I from point G [by book 5, prop. 17]. So draw line IO. It is also clear [from book 5, prop. 7] that AG:AN = GI:IN, while BG:BM = GO:OM. Thus, since lines AG and BG are each cut in three points according to this ratio,
and since two of the lines, i.e., AB and MN, extended from the points of division [A and N on base AG] intersect at the same point, i.e., at the same point Q, the third [line extending from point I on base AG] will necessarily intersect [these two] at that same point [by proposition 6, lemma 3].

4.120◉ Therefore, when it is extended, IO will fall upon Q, so IOQ [forms] a straight line. Thus, IOU will not be a straight [line]. But IOU is the image of line AQ, so the image of line AQ will be curved. Furthermore, if point B replaces point Q, and if some point on line AB replaces point B, it will be demonstrable in exactly the same way that the image of line AB is curved, and this
is what was set out [to be proven].⁑

4.121◉ [PROPOSITION 13] If, however, AB intersects the circle [in figure 6.4.13, p. 116], let it intersect at point E, [and let] M [be] the endpoint of tangency on line BG. [The form of point] B is reflected to [point] D from some point [F’] on arc HP. The arc [extending] from that point of reflection to H [i.e., HF’] is either equal to, longer than, or shorter than arc HE.

4.122◉ If it is equal (but it is evident that that arc is equal to arc HQ), let Q [in figure 6.4.13] be the point on the circle where the tangent drawn from point M falls on the side of E. Thus, AE passes through point Q, so MQ intersects AE through point E.⁑

4.125◉ Whether the latter or the former is the case, repeat the earlier proof, and it should be demonstrated in precisely the same way that the image of line AB is curved, which is what was set out [to be proven].

4.126◉ [PROPOSITION 14] Furthermore, if the center of sight lies in the plane containing the visible line and the center of the sphere—in the previous cases it was stipulated that the center of sight does not lie in that plane⁑—then the straight visible line will either intersect the [great] circle [forming] the common section of that plane and the mirror, or it will not intersect [it].

4.127◉ If it will intersect, it will be either perpendicular to the mirror[‘s surface] or inclined to it. If [it is] perpendicular, the angle [formed by] those lines will fall on the center of the mirror, and [the image of] that line will appear straight, for the image of any point on that line will appear on that line, and so the image of that line [will be] straight.

4.128◉ But if the given line [AB] is slanted, its slant will either be toward the center of sight or away from it. If it slants away from it [as in figure 6.4.14, p. 120],⁑ find the point [R] on the circle from which [the form of] some [point on it, such as B’] is reflected to the center of sight [D], and find the line of reflection [RD]. Any of the slanted lines may fall on this line of reflection, and if it does, then [an image of] this slanted line will not be seen.

4.129◉ Having extended a line [DG] from the center of sight to the center of the mirror, take a point [R’] on the arc of the circle in front of this line, such that [the form of] some point [B’] on the slanted line is reflected from it to the center of sight. But [the form of] that point is reflected from the previously designated point [R], which is the endpoint of the line of
reflection, since the slanted line lies upon the line of reflection, and so [the form of] that point on the slanted line is reflected to the center of sight from two points on the arc, which is impossible [by book 5, prop. 16].

4.130◉ Moreover, even though [the form of] that point may be reflected from the point that is initially selected, it[s image] is still not seen, since it lies on the line of reflection, so it[s image] is occluded by points in front of it [on the object-line], and so [the image of] a line lying upon the line of reflection is not seen.⁑

4.131◉ But if a slanted line [AB in figure 6.4.14a, p. 120] is taken with its slant not toward the center of sight, and if it lies below the line of reflection [AB] and cuts it at a point [B] on the circle, I say that no [image of any] point on that line will be seen.

4.132◉ For, given [such a] point [e.g., A], if it is claimed that [the form of] that point can be reflected from some point [R’] on the arc lying between the line of reflection [DB] and line [DG] extended from the center of sight to the center of the mirror, and if a line [of incidence AR’] is extended from that point to the point chosen on the arc, this line will intersect the line of
reflection [BD], and the point of intersection [X] is reflected to the center of sight from two points on the arc, which is impossible.⁑

4.133◉ On the other hand, if it is claimed that the [form of the] point [A] taken on the [slanted] line is reflected from a point on the arc of the circle below that line [i.e., to the right of B], it will be impossible [for it to be seen], since that whole arc is occluded by the line.

4.134◉ If, however, the chosen line does not reach the circle, it can indeed be seen, but it is quite small. But if a line with the previous slant [i.e., away from the eye] is selected between the line of reflection and the line initially assumed to pass through the point of reflection to the center [of the circle], this line can in fact be seen, and the curvature of the image of this
line will decrease as it approaches the line passing through the point of reflection to the center [of the circle].⁑

4.135◉ But if lines are chosen between the line passing through the point of reflection to the center [of the circle and the mirror], they will appear, whether their slant lies toward the center of sight or not. And the way they [i.e., lines slanting toward the eye] are seen is like the way lines [slanting away from the eye] between the line of reflection and the line passing to the
center are seen. But these things must be understood for lines that meet the arc of the visible part of the circle, i.e., on the arc lying between the two [lines] drawn tangent to the circle from the center of sight.

4.136◉ On the other hand, among lines that meet the circle on the side of the circle that is invisible, one of them will be parallel to the line of reflection. That one will not be seen. Likewise, any one that borders on the parallel and lies below it will be invisible, whereas one that borders on the parallel [and lies] above it can be seen.⁑

4.137◉ If, however, a line is selected between the parallels but not bordering on any of them, and if it is slanted toward the center of sight, it will be seen. If it slants in the other direction, it will sometimes be seen, and sometimes not because, if a parallel to the line of reflection is extended from its endpoint, and if that line lies below the parallel, then it will not be seen,
whereas [if it lies] above it can be seen.⁑

4.138◉ But if the lines do not meet the circle, they will either intersect the line drawn from the center of sight to the center of the mirror, or they will be parallel to it. If any of them intersects it, that line will either intersect it on the side of the center of sight, i.e., between the center of sight and the mirror, or [it will intersect it] beyond the mirror [and the center of
sight]. If [it intersects] beyond [i.e., above the head], that line will not be visible, but its ends may appear. If it cuts the visual axis on the side of the center of sight, it will in fact appear the same [i.e., not visible in the mirror]. If it is parallel to the visual axis, it can be seen.⁑ Moreover, the images of all these lines are curved.

4.139◉ And if the center of sight lies in the same plane as the center of the mirror and the visible lines, they appear diminished, and the one that appears most clear in this case is the one that is most slanted and that corresponds to the center of sight. By the same token, the images of arcs that appear in these [sorts of] mirrors and that lie in the same plane as the center of the
mirror and the center of sight appear curved according to the curvature of the mirror.⁑

4.142◉ Let D be the center of sight and G the center of the mirror. Let HE be the visible line. Let HE not intersect the circle but be parallel to line DG [as in figure 6.4.15, p. 122], or let it intersect it on the side of D [as in figure 6.4.15a, p. 122]. Take the plane containing line DG and line HE, and let circle AB be the common section of this plane and the mirror.

4.143◉ Draw line HG. Let Z be the image of H, let B be the point on the circle from which [the form of point] H is reflected to [point] D, let the tangent be drawn from point B, and let it intersect line HG at point T. T will be the endpoint of tangency [on cathetus HG].

4.144◉ Draw line GB, which will necessarily intersect HE when it is extended, for, if HE is parallel to DG, it will necessarily intersect. If, how-ever, DG intersects HE, then a fortiori GB will intersect it. That intersection will lie either on line HE, or beyond that line.

4.145◉ Let it lie beyond. Let it intersect at point M, let Q be the image of point M, and let S be the endpoint of tangency [on cathetus MG]. Draw line ZQ, as well as line TS, and draw tangent AU from point A. It is clear that [arc] AB is less than one-fourth [the circumference of the circle, since GD lies on a diameter of the circle and DB intersects the circle], so [the eye at point] D
should see less than half the circle [when the corresponding arc below A is included], [and] so angle AGB is acute, while angle UAG is right. Hence, AU will intersect GB. Let it intersect at point U. I say that point U should fall above point S.

4.146◉ For, since [the form of] point M is reflected from some point [X] on arc AB, and since A lies below that point, the endpoint of tangency for A [as a point of reflection for the form of any point on cathetus GM] will lie higher than the endpoint of tangency for that point [X as a point of reflection for any point on cathetus GM]. And so S [lies] below point U. Accordingly, extend
TS until it intersects line AU, and let the intersection be at point K.

4.147◉ Draw line GK, and let it intersect HM at point C when it is extended. [The form of] point C is reflected to [point] D from some point on arc AB. Let F be that point, and from it draw a tangent to GC, that tangent being lower than line AK, and [any] point [on it, such as] O will be lower than point K.

4.149◉ For either lines HC, TK, and ZL are parallel, or they will intersect. Let them be parallel [as in figure 6.4.15a]. Accordingly, since they are parallel, let them intersect line CG at the three points C, K, and L, and let them intersect both lines MG and HG. HG:HT = GZ:ZT [by book 5, prop. 7], and likewise MG:MS = GQ:QS [because HG and GM are cut equiproportionally by parallels
HMC, TSK, and ZQL, and for that same reason] GC:CK = LG:LK [all according to proposition 7, lemma 4].

4.150◉ But it is clear that R is the image of C because line of reflection DF intersects CG at point R, and O is the endpoint of tangency, so GC:CO = GR:RO [by book 5, prop. 7]. However, GC:CK [which = GL:LK] > GC:CO [which = GR:RO], and so GL:LK > GR:RO. Accordingly, OR:RG > KL:LG, and so OG:RG > KG:LG [by Euclid, V.18]. But KG > OG, so LG > RG. Hence, R lies lower
than point L. But ZQL is a straight line. Therefore, ZQR is a curved line, and so the image of line HC is curved. So if some point on line HC replaces point M, and if point E replaces point C, it will be demonstrable that the image of HE is curved.

4.151◉ But if lines HC, TS, and ZQ intersect, the intersection will either be on the side of D, or [it will be] on the side of HG. Let it be on the side of D [as represented in figure 6.4.15b, p. 123], and let the intersection be at point C. ZQC will be a straight line, so ZQR will be curved, and so the image of line HE [will be] curved, which is what was set out [to be proven].⁑

4.152◉ If an arc is posed outside the mirror, however, it will be possible from this to prove that its image is curved just as it was proven [in proposition 11] when the center of sight did not lie in the same plane as the arc and the center of the mirror, and this is what was set out [to be proven].

4.153◉ Therefore in these [sorts of] mirrors straight lines appear curved, and likewise curved lines appear curved. Moreover, if a curved object is placed before the eye in these mirrors, and if it is long and has some slight breadth, the curvature of that object will appear clearly, since it can be detected by those features lying on or within the body. In fact, unless it is
considerable, the curvature is not clearly detected when the boundaries of the length or breadth are hidden [so that the image extends beyond the visible face of the mirror], so when an object of slight curvature and considerable size is placed before the eye, its curvature is not clearly detected, even though its image is curved, since the boundaries along the length and breadth of the object do not appear [in the mirror].

4.154◉ Moreover, all of the errors that occur in plane mirrors occur in these mirrors as well,⁑ and besides those [errors] it happens that the images of straight lines are curved, which is far from the case in plane mirrors.

5.1◉ Now the same errors occur in convex cylindrical mirrors as occur in convex spherical mirrors, for [in the former] as in the latter, straight lines appear curved and the size of the visible object appears diminished, but far more pronouncedly because in [convex] spherical mirrors a large object will appear smaller, to be sure, but not very small, whereas in convex cylindrical ones even
a very large object will appear greatly diminished [in size]. Likewise, a straight line will appear curved in [convex] spherical mirrors, but if it is [even] slightly curved [it will appear] extremely so in [convex] cylindrical [mirrors], so the errors [that occur] in a [convex] spherical mirror are compounded in a [convex] cylindrical mirror.

5.2◉ However, in [convex] cylindrical [mirrors] reflection sometimes occurs from a straight line, i.e., [when it occurs] from [a line of] longitude on the mirror, sometimes from a circle [on the mirror’s surface], and sometimes from a [cylindric] section [i.e., an ellipse, on that surface]. When the visible line is parallel to a [line of] longitude on the mirror, the reflection will occur
from [that] line of longitude, and a straight visible line will appear [only] slightly curved. These things, moreover, will be demonstrated, but for that demonstration a preliminary point must be made, as follows.

5.3◉ [PROPOSITION 16, LEMMA 5] If a cylindric section [e.g., ellip-tical section APEBR in figure 6.5.16, p. 124] is assumed and some point [E] is taken on it that is not a point of reflection, then, when a line [ED] is extended from that point to the normal [BD dropped] from the point of reflection [B] to the axis [of the cylinder on which the ellipse is chosen]—and that line should form
an acute angle [EDB] with the normal—if a line [EU] is drawn perpendicular to the tangent [QEL] at that point [E], this line will intersect the normal [BD] outside the axis and outside the intersection of the previous line [ED] with the normal [BD].⁑

5.4◉ For example, let AEB be the [assumed cylindric] section, E the given point [that is not the point of reflection], N the visible [object-]point, B the point of reflection, BD the [given] normal [dropped from the point of reflection to the axis], EDB an acute angle, and QEL the tangent [to the cylinder as well as to the elliptical section at the chosen point E].

5.5◉ At point B form a circle, i.e., BTO, parallel to the cylinder’s base, and draw a line of longitude through point E on the cylinder, i.e., ET. Draw axis DH [of the cylinder], and draw line DC perpendicular to BD.

5.6◉ It is obvious that plane HDC is orthogonal to the plane of the circle [BTO]. But the plane tangent to the cylinder at point B will be parallel to this plane [HDC] because the line of longitude extended from point B will be parallel to the axis, and the tangent at [point] B [along the line of longitude dropped through it] will be parallel to CD. Therefore, the plane containing lines LE
and ET is not parallel to plane HDC. It will therefore intersect that [plane HDC]. Let it intersect along line LC, and draw line TC, which will of course be tangent [to the cylinder], since plane LET is tangent [to it, by construction]. Moreover, when line TD is drawn, angle CTD will be right because TD is a [a radial segment of the] diameter [of circle BTO, and CT is tangent to the circle at its endpoint].

5.7◉ Now at point E form a circle, i.e., ESP, on the cylinder parallel to the base. Let K be the point on the [cylinder’s] axis [where it intersects] this circle, and draw line KE. Draw line DL, as well, and it will certainly intersect the plane of circle ESP. Let it intersect at point F, wherever that point may lie either outside or inside the circle, and draw lines KF and EF.⁑ Then from point
F draw FM perpendicular to the plane of circle BTO, and draw line TM.

5.8◉ It is evident that KD is parallel and equal to FM [since they are perpendicular to parallel planes], and so KF is parallel and equal to DM. Likewise, KD is parallel and equal to ET, and KE is parallel and equal to DT. Hence, TE will be parallel and equal to FM, and so EF [will be] parallel and equal to TM.

5.9◉ But plane KDL is perpendicular to plane BEO of the [cylindric] section, and it is perpendicular to the plane of circle ESP. Therefore, it is perpendicular to common section EF of the [cylindric] section and the circle. So angle EFK is right. Likewise, angle TMD is right [since DM and KF are parallel, as are TM and EF].

5.10◉ Therefore, since angle DTC is right [by construction], rectangle DM,MC = rectangle TM,FE,⁑ but since FM
is parallel to CL [because CL is necessarily parallel to TE, given that it is the common section of planes TCLE and HDT, which are both perpendicular to the plane of circle ESP], then [triangles DFM and DLC will be similar and will have corresponding sides proportional (by Euclid, VI.4), from which it follows that] DF:FL = DM:MC. But DF > DM, so FL > MC. Consequently, rectangle DF,FL > rectangle DM,MC, so, since TM = EF, rectangle DF,FL > rectangle
EF,FE [which = rectangle TM,FE, which = rectangle DM,MC], so angle LED > a right angle, for if it were a right angle, then because line EF is perpendicular to LD, rectangle DF,FL would be equal to EF2.⁑ It therefore follows that angle DEQ [adjacent to angle LED] is acute. Hence, the perpendicular [EU] dropped from point E, that perpendicular being perpendicular to tangent QL, will fall outside line ED and will intersect normal BD outside point D, which is what was set out [to be proven].⁑

5.12◉ [PROPOSITION 17] Assume a cylinder [in figure 6.5.17, p. 126], and let TH be a [visible] line parallel to the [cylinder’s] axis [ZK]. TH will of course be parallel to the line of longitude [AG in the same plane with TH and the axis] of the cylinder.

5.13◉ Therefore, if the center of sight [E] lies in the same plane as the axis and line TH, the [form of the] line can be reflected, and the reflection will occur from the line of longitude on the cylinder, which is the common section of the plane containing the center of sight and the axis and the surface of the cylinder, as was shown in [proposition 28 of] book 5. Line TH will thus
appear as a straight line [T’H’] because any normal dropped from a point on line TH [such as TT’ or HH’] will lie in the same plane as the center of sight and the axis, and it will be proven that the image of line TH is straight, just as it is proven for [straight] lines seen in plane mirrors.

5.14◉ Let the center of sight [E in figure 6.5.17a, p, 126] lie outside the plane containing line TH and the axis, and [let] TH be parallel to the axis, which is ZK. Project a plane that passes through the center of sight and cuts the cylinder’s surface parallel to the base. It will of course cut a circle [on that surface]. Let that circle be BF. [The form of] some point on line HT is
reflected to the center of sight from some point on this circle. Let it be [reflected] from point B, and let E be the center of sight.

5.15◉ Let Q be the point on line TH [whose form is reflected to E from B], draw lines EB and QB, draw line of longitude ABG from point B, and draw the normal ML through point B that intersects the axis at point L. Then from point E extend line EO parallel to ML, and extend QB until it intersects [EO]. Let the intersection be at point O.

5.17◉ Now choose another point on line TH, let it be point T, and draw line TO. It is clear that line TH is parallel to line of longitude AG [by construction]. Therefore, they lie in the same plane, and line QBO lies in that plane, so line TO will lie in [that] same [plane]. Hence it will intersect line AG. Let it intersect at point G, and draw line EG.

5.18◉ It is also clear that line AG is perpendicular to the plane of circle BF, as is the axis [ZK] to which it is parallel, and its plane [is] EOBF [which] cuts the cylinder parallel to its base. Thus, angle GBO [is] right, and [so] angle GBE is right. Consequently GO2 = GB2 + BO2 [by Euclid, I.47]. Likewise, GE2 = GB2 + BE2, and since BE and BO are equal [by previous conclusions], while
GB is common, GO = GE. Hence, angle GOE = angle GEO [in isosceles triangle GEO].

5.19◉ Moreover, if normal ZGN is drawn, it will be parallel to EO, since it is parallel to MBL [to which EO was made parallel by construction]. Therefore, angle TGN = [alternate] angle GOE, and angle NGE = [alternate] angle GEO, so angle TGN = angle NGE. Furthermore, since E, O, N, G, and Z lie in the same plane, and since G lies in that plane, E, G, and T will lie in the same plane, and
so lines EG, NG, and TG lie in the same plane [which is therefore the plane of reflection]. Thus, [the form of] T is reflected to E from point G.

5.20◉ Furthermore, if point H is taken on line TH at the same distance from point Q as point T, and if line HO is drawn, it will pass through [some] point on line AG. Let it pass through point A. When normal DA[Z’] and lines EA and HAO are drawn, it will be a matter of proving as before that the two angles ABO and ABE are right, that the two sides AO and AE are equal, and that the two
angles HAZ’ and EAZ’ are equal. And so [the form of point] H is reflected to [point] E from point A. Likewise, if any [other] point on line TH is chosen, it will be a matter of proving that [its form] is reflected to E from another point on line AG, so [the whole form of] line TH is reflected from line of longitude AG.

5.21◉ [PROPOSITION 18] It remains to demonstrate that the image of line TH is curved. It is clear from the preceding [theorem] that [the form of] Q [in figure 6.5.18, p. 127] is reflected to E from point B, which is a point on circle [FB]. But since it is reflected in this way from the circle, if a line is drawn from point Q to the center of that circle, it will meet the normal [MBL]
dropped from point B, and the intersection [of these two lines] will lie at a point on the axis. So draw QL intersecting ML at point L on the axis, and [this] is the center of circle FB. Then extend EB until it meets QL. Let the intersection be at point C. C will be the image of Q, and C lies in the same plane with lines QH and the axis [ZK], and [with] line of longitude AG.⁑

5.22◉ It is also evident that [the form of] T is reflected to E from a point on a [cylindric] section of the cylinder, namely, point G. Moreover, [as established in proposition 16, lemma 5] a line can be drawn from point T perpendicular to a line tangent to another point on the [cylindric] section, and it will intersect normal NGZ dropped from point G outside the axis, that is, outside
point Z, which is the intersection of normal NZ and the axis, for if line TZ is drawn, angle TZN will be acute [as stipulated in proposition 16]. Accordingly, draw TX [normal to the cylindric section, as prescribed, and] intersecting NZ at point X, and extend EG until it intersects TX at point I. I will be the image of point T.

5.23◉ Likewise, when the line [HP] orthogonal at a point on the [cylin-dric] section from which reflection [of the form of point H] occurs is drawn from H, it will intersect normal DAZ’ outside of point D, which is a point on the axis. Let it intersect at point P, and extend EA until it intersects HP at point S. Point S will be the image of point H. Now draw line SI.

5.24◉ It is clear that, since line TI intersects normal NZ, which is parallel to line EO, it will intersect line EO. The same holds for line HS; because it intersects normal DAZ, which is parallel to EO, it will intersect EO. But since T’s location with respect to point E is equivalent to and the same distance [from E] as H’s location [by construction], the location of point T and of
point H [will] likewise [be] equivalent with respect to point O, and [that of] points I and S is also equivalent with respect to O. The location of lines TI and HS will also be equivalent with respect to line EO.⁑

5.25◉ Lines TI and HS will therefore intersect at the same point on line EO [since each lies in a plane with it, and both are inclined toward one another]. Let them intersect at point U. TUH will [therefore] be a triangle, and [straight] line IS will lie in the plane of this triangle. The axis, however, does not lie in this plane [since normals TIU and HSU bypass it].

5.26◉ But TH does lie in the same plane as the axis so that plane [TZKH] intersects the plane of the triangle [TUH] along common section TH, not along some other [line of section]. Therefore, since point C lies in the plane of line TH and the axis, but not on line TH [itself], it does not lie in the plane of triangle TUH, whereas the two points I and S do lie in the plane of that
triangle, so line ICS is curved, and the image of line TH will [therefore] be curved, which is what was set out [to be proven].

5.27◉ But its curvature is slight because the perpendicular dropped from point C to the plane of the circle is extremely small,⁑ and the closer the visible line is to being parallel to the line of longitude on the mirror, the less sharply curved it[s image] is, [whereas] the farther [it is from such parallelism] the more [sharply curved its image is].⁑

5.28◉ [PROPOSITION 19] Furthermore, if line TH intersects the plane containing the center of sight and the axis, and if it is orthogonal to it, the center of sight will either lie in the plane of line TH intersecting the plane of the axis and the center of sight orthogonally, or [it will lie] outside [that plane].

5.29◉ If [the center of sight] lies in that plane, it will lie beyond or in front of line TH. If [it lies] beyond, then, since that line has bodily dimensions, it will block the mirror from the center of sight, and so it[s form] will not be reflected [to the eye], although perhaps its terminal segments will appear and be reflected from the circle on the cylinder that forms the common
section of the plane of line TH that cuts the cylinder and the cylinder [itself]. And the image of these terminal segments will [appear] just as [they do] in convex spherical [mirrors, as described in proposition 14, paragraph 4.138].⁑

5.30◉ Likewise, if the center of sight lies in front of TH, part of that line will be hidden by the head containing the center of sight. Nonetheless, the visible part of the line is reflected [to the center of sight] from the circle [formed by the plane of reflection] in exactly the same way as in convex spherical [mirrors, according to proposition 14, paragraph 4.138].

5.31◉ But if the center of sight lies outside the plane of TH that cuts the plane of the center of sight and the axis orthogonally, then let E [in figure 6.5.19, p. 128] be the center of sight [above line TH] and XZG the cylinder.⁑ [The form of point] H is reflected to E from some point on the cylinder. Let [it be reflected] from B. Let T lie the same distance [as H] from point E. I
say that [the form of point] T is reflected to E from another point [i.e., other than B] on the cylinder, and that, since points H and T are equivalently situated and the same distance from point E, their points of reflection, i.e., B and G, will be equivalently situated and the same distance from point E. Therefore, the two points B and G will lie on a circle.

5.32◉ Let BZG be the circle, with D its centerpoint. Draw lines HB, BE, TG, and GE, and from the centerpoint [D] draw normals to the tangents at B and G, i.e., [normals] DBO and DGS. Then draw line ED, and extend HB and TG until they intersect line ED.

5.33◉ Since points H and T are equivalently situated and the same distance with respect to E and with respect to D, and since, by the same token, points B and G are equivalently situated with respect to D and with respect to E, lines HB and TG will be equivalently situated with respect to line ED, and so they will intersect at the same point on that line. Let [that intersection] be at
point L.

5.34◉ Produce the line of longitude on the cylinder that contains point Z, let this line lie in the plane containing the center of sight and the axis, let it be AZ, and draw [lines] LZN and DZC. Let Q be a point on line TH, that is, the point [on it] that lies in the plane of the center of sight and the axis, and from point Q draw a line parallel to line DZC. This line will fall on the
axis, and LZN will fall on this line beyond point Q. Let it fall at point N.

5.38◉ Let [angle FCZ] equal [to angle CZE] be cut [from NCZ] by line FZ, which will intersect line NQ [at point F] beyond point N. Therefore, since angle FZC = angle CZE, [the form of] F is reflected to E from point Z. [The form of point] Q is reflected to E from a point on the line of longitude passing through Z, that is, from a point on AZ beyond [i.e., below] Z. For if [it occurs] from
a point this side of Z, i.e., nearer E, the line extended from point Q to that point of reflection will cut line FZ, and so the point of intersection is reflected to E from two points, which is impossible.⁑

5.39◉ Take point K below Z from which [the form of] Q is reflected to E, then, and extend line EK until it intersects line NQ [i.e., the cathetus dropped from object-point Q] at point P. P will be the image of Q. But [the form of] H is reflected to E from a point on the cylindric section [formed on the cylinder by plane of reflection HBE]. Therefore, if a normal is dropped from point H to
the line tangent to the [cylindric] section at some point [on it], that normal [i.e., the cathetus] will intersect normal CZD outside the axis [by proposition 16, lemma 5]. Let it intersect at point U.

5.40◉ Likewise, from point T a normal can be drawn to the [cylindric] section from a point on which [its form] is reflected to E. And since points H and T are equivalently situated with respect to line CZD, the same also holds for the points on the [cylindric] section through which the normals [i.e., the catheti] pass, so those two normals will intersect at the same point on line CZD.
Accordingly, let them intersect at point U.

5.41◉ [Therefore, the extension of] line EB will intersect line HU. Let R be the point of intersection. By the same token, let EG intersect TU at point Y, and draw line RY.⁑ It is obvious that R is the image of H, [and] Y is the image of T, and we have triangle ERY.
Point Z lies outside the plane of this triangle, so the plane of this triangle is higher than line EP, and so P lies outside [that plane]. Hence, line RPY will be curved, and it is the image of line TH, and the curvature of this image is certainly not inconsiderable, which is what was set out [to be proven].

5.42◉ It is therefore clear that in these [sorts of] mirrors, if a straight visible line is parallel to a line of longitude on the cylinder, its image will be either straight or verging toward straightness. But if a straight visible line is parallel to the width of the mirror [i.e., the plane through it is perpendicular to the cylinder’s axis], its image will be curved, and its curvature
will not be inconsiderable.

5.43◉ Furthermore, among [visible] lines oriented between these two [extremes], the images of those that verge more closely toward an orientation parallel to the longitude of the cylinder will be closer to straight, whereas the images of those that are nearer to an orientation parallel to the [cylinder’s] width will be more curved. And the curvature of the images will diminish or augment
depending on how close or far the lines are from either of these orientations, and this is what was set out [to be demonstrated].

6.1◉ Moreover, in convex conical mirrors the same errors occur as happen in convex cylindrical [mirrors],⁑ for
[straight] visible lines that are parallel to the longitude of the cone appear straight, whereas those parallel to the width [of the cone appear] curved, and for those at intermediate positions, their curvature augments or diminishes according to how near or how far [they are from those extreme positions], and this will of course be demonstrated. However, we must set forth something beforehand, and it is [as follows].

6.2◉ [PROPOSITION 20, LEMMA 6] If a point of reflection is taken on the surface of a cone and a [conic] section is produced [on that surface] to pass through that point, and if a point is taken on that [conic] section farther from the vertex of the cone than the point of reflection and a normal is dropped from the selected point to a line tangent to the [conic] section, this normal will
intersect the normal dropped from the point of reflection [at a point] outside the axis.

6.3◉ For instance, let ABGZ [in figure 6.6.20, p. 129] be a cone standing upright on its bases [i.e., a right cone], A the cone’s vertex, BFZ the [conic] section [produced on its surface], E the point of reflection, and Z the point on the [conic] section farther from [vertex-]point A than E. At point Z let there be a plane cutting the cone parallel to its base. It will of course cut it
along a circle [forming the] common [section of the cutting plane and the cone’s surface]. Let that circle be GBRZ, draw lines AZ and AE, and extend AE until it is equal to AZ. It will reach the circle. So let it fall at point O on it.

6.4◉ Let C be the center of the circle, draw axis AC, and from point E draw the normal [ED] to the plane tangent to the cone [at that point]. It will of course intersect the axis in the vicinity of the circle’s centerpoint C. Let [that intersection] be at point D, and draw line DZ.⁑

6.5◉ Then from point O draw a normal [OK] intersecting the axis at point K, and draw lines DZ and KZ. At point Z produce [line] TQ tangent to the [conic] section, and [at the same point produce] another [line] ZY tangent to circle BGZ.

6.6◉ Next draw line BCZ, and from point C draw CR perpendicular to line BCZ. It will of course be perpendicular to the axis, since the axis is perpendicular to the circle [in whose] plane [CR lies], so CR is perpendicular to plane ACZ. It will also be parallel to tangent ZY, so ZY is perpendicular to plane ACZ, [and] so TQ is not perpendicular to that same plane.

6.7◉ However, because K is the [the endpoint of] pole [KC] in circle BRZ, then, since lines KO and KZ are equal [because they are lines of longitude on a right cone with its vertex at K and its base circle passing through Z and O], and since axis AK is common [to triangles AOK and AZK], angle AOK = angle AZK, and so angle AZK is right [since angle AOK was constructed as a right angle].
Therefore, since line KZ is perpendicular to [line] AZ, which is a line of longitude [on the mirror], it will be perpendicular to the plane tangent to the cone[’s surface] along this line of longitude. But TQ lies in the [same] tangent plane because it is the common [section] of the tangent plane and the [conic] section. Accordingly, KZ is perpendicular to TQ.

6.8◉ Furthermore, draw HZ in the plane of the [conic] section perpen-dicular to line TQ [and therefore normal to the section itself]. Since line KZ lies outside the plane of the [conic] section, it will intersect line HZ and will [therefore] not form a single line with it. Hence, plane KZH intersects the plane of the [conic] section along common section HZ, and it intersects line TQ at
point Z. In addition, plane AZK intersects plane AZH along common section KZ.

6.9◉ But DZ lies in the plane of the [conic] section, and it is intersected by line KZ at point Z, and point T [lies] above plane KZH, point Q below [it, i.e., on the other side of it from T]. And so plane KZH cuts plane DZQ along a common section, and that common section is perpendicular to line TQ because that line lies in plane HZK to which TQ is perpendicular. And since plane HZK
intersects plane DZQ, and since plane HZK slants in the direction of [segment] ZE [of the conic section], the common section [HZX] of those planes will lie between lines QZ and DZ, and so it will intersect normal ED [at point X] outside the axis. That it [i.e., the normal to the section at Z] must necessarily intersect it [i.e., normal ED dropped from center of sight E] has been demonstrated in book 5, proposition 26,84 and so what was set out [to be demonstrated
has been shown].⁑

6.10◉ [PROPOSITION 21] So let there be a [right] cone [in figure 6.6.21, p. 129] with its vertex at A, AH being its axis and AZ a line of longitude, and from point Z to the plane tangent to the cone along line AZ drop a perpendicular, which will perforce intersect the axis [at point H]. Let it be line TZH.

6.11◉ From point A extend line AN outside the cone [and] above the plane tangent to the cone along line AZ so that it forms an acute angle with the axis, as well as with line AZ of longitude. From point H within plane AHN, draw line HO forming an angle [AHO] equal to angle AHZ, that line necessarily intersecting line AN [at point O. Consequently] when a circle is produced through point Z
parallel to the [cone’s] base, HO will pass through [that] circle just as HZ passes through it.

6.12◉ Now draw line OZ, and extend it to point F. Since line OZ lies above the plane tangent to the cone along line AZ, then because HZ is perpendicular to that plane [by construction], angle OZH will be greater than a right angle [because it intersects the plane tangent to AZ from above]. Consequently, [adjacent] angle FZH is acute [so that ZF lies inside the cone].

6.13◉ From point Z draw ZM tangent to the circle, and from point F draw a line perpendicular to AZ to fall at point E on it, and when it is extended, it will intersect AO because angle OAZ is acute [by construction, and because AO, AZ, and OZF lie in the same plane]. Accordingly, let it fall at point N, and from point E draw line QE parallel to line TH.

6.14◉ Then from point E draw LE parallel to line MZ. It is evident that MZ is perpendicular to AE because it is perpendicular to TH, as well as to the diameter of the circle, to which it is tangent. Therefore, LE is perpendicular to AE [since it is parallel to MZ, by construction].

6.15◉ Now produce plane LQD cutting the cone. It will of course form a conic section [because the cutting plane intersects the axis below the circle passing through E]. Hence, since AE is perpendicular to FN [by construction], as well as to QD and LE, FN will lie in that plane [QEDL], which cuts the cone [along the aforementioned conic section]. Accordingly, produce CF parallel to QE. It
will of course be parallel to TZ.

6.16◉ But since angle OZT is acute [by previous conclusions, adjacent] angle TZF is obtuse. From point Z draw a line [ZC] that forms with TZ an angle [TZC] equal to angle OZT, and that line will necessarily intersect FC. Let it intersect at point C, and draw line EC. Therefore, since CZ and OZ lie in the same plane, and since angle OZT = angle TZC [by construction, the form of] point O is
reflected to C from point Z.

6.17◉ Since, however, angle OZT = [alternate] angle ZFC, and since angle OZT = [alternate] angle ZCF, sides ZC and ZF [of triangle ZCF] will be equal, and given that angle FEZ is right [by construction], FZ2 = EZ2 + EF2 [by Euclid, I.47], while CZ2 = EZ2 + EC2 [by the same theorem]. Therefore, CE and FE are equal, and so angles ECF and EFC are equal, whereby angles NEQ [which is alternate
to EFC] and QEC [which is alternate to ECF] are equal. And since C, E, and N lie in the same plane [i.e., the plane producing the conic section through point E, the form of] point N is reflected to C from point E.

6.18◉ Likewise, draw some line from point F to some point on line ZE, and extend it to ON. As to the point on line ON where it falls, it will be proven that [its form] is reflected to C from the [corresponding] point on ZE because it cuts that line. In the same way, as well, for all such lines, the proof will take its start from perpendicular FE with respect to line EZ, which will be the
common terminal [base-line], and so [the form of] any given point on line ON is reflected to C from some point on line EZ.

6.19◉ [PROPOSITION 22] Having demonstrated this point, we should state [that], when the eye perceives straight lines that pass through the vertex of a right convex conical mirror at a slant to the mirror’s axis, the forms of those [lines] will be somewhat convex in that mirror.

6.20◉ Accordingly, let the right conical mirror be ABG [in figure 6.6.22, p. 130], with A as its vertex and AD as its axis, let us produce line AZ at random on its surface, and let point Z be marked on it at random. Let a plane pass through Z parallel to the base of the cone, and let it form circle ZU. Then from Z let us drop ZH perpendicular to AZ. Hence, this line will intersect the
cone’s axis, so let it intersect [that axis] at H.

6.21◉ From Z let us extend line ZM tangent to the circle [ZU], and let us extend a line from A that forms an acute angle with both lines AZ and AH, and let it lie outside the plane tangent to the cone that passes along line AZ, which is possible. Let it be AO, then, and let us produce a line from point H within the plane containing AO and AH that forms an angle [AHO] with AH equal to
angle ZHA. This line [HO] will therefore intersect AO because the two angles at A and H [i.e., OAH and AHO] are acute. So let them [i.e., AO and HO] intersect at O.

6.22◉ Accordingly, line HO will intersect the circumference of circle ZU because angle AHO = angle AHZ [by construction]. So let it intersect at U, and let us extend AU in a straight line. Let us also extend perpendicular HZ to T, let us produce OZ and extend it directly to F, and extend AZ to E. Therefore, angle FZH will be acute because line OZ cuts the plane tangent to the cone and
passing along AZ. Hence, line FZ lies below the common section of plane OZH and the [aforementioned] tangent plane [passing along AZ], and this common section forms a right angle with line HZ. Thus, angle OZH is obtuse, [and] so [adjacent] angle FZH is acute.

6.23◉ So take point F on ZF, from it extend FE perpendicular to AE, and continue it in a straight line. It will therefore intersect line AO because angle OAE is acute. Let it intersect at N, then, and extend line ED from E parallel to line ZH. ED will thus be perpendicular to the plane tangent to the cone passing along AE.

6.24◉ Then from E draw line EL parallel to line ZM, and produce the plane containing LE and ED. It will therefore intersect the surface of the cone and will form a [conic] section, for this plane is oblique to axis AD.

6.25◉ Let the [conic] section be BEG’. MZ is perpendicular to plane AZH [since it was constructed tangent to the circle at point Z and is therefore perpendicular to the diameter passing from Z through axis AH of the cone], and this was established earlier [in proposition 21]. Therefore, line LE is perpendicular to plane AED, so angle AEL is right, angle AEN is right, and likewise angle
AED is right [all three by construction]. Consequently, lines LE, NE, and DE lie in the same plane [to which AE is perpendicular]. Hence, line FEN lies in the plane of the [conic] section [BEG’].

6.26◉ From point F extend line FR parallel to line DE. Accordingly, this line will be parallel to line HZ [to which DE was constructed parallel]. In plane OZH draw a line [ZR] from Z that forms with ZT an angle equal to OZT. This line will therefore intersect FR because it will intersect ZH, which is parallel to FR [by construction] and lies in the same plane with it, since ZF lies in
that plane. So let it intersect at R.

6.27◉ Accordingly, the two angles at R and F [i.e., ZRF and ZFR] are equal, for they are equal to the two angles [OZT and TZR] at Z.⁑ So the two lines RZ and FZ [within
isosceles triangle ZRF] are equal. But it has been shown that line FEN lies in the plane of the [conic] section [BEG’], and line FR is parallel to line ED [by construction]. It [i.e., FR] therefore lies in the plane of the [conic] section.

6.28◉ Let us then draw RE. It will thus lie in the plane of the [conic] section. Extend DE to K, and it has been shown that EA is perpendicular to the plane of [that] section. Hence, each of the angles AER and AEF is right, and the two lines FZ and RZ are equal. Consequently, the two lines RE and FE are equal, so the two angles ERF and EFR are equal.

6.29◉ Accordingly, the form of N will be reflected to R from E, and the form of O will be reflected to R from Z. Moreover, every line extended from F to some point on line AN will intersect AE. But it is clear that that line will be equal to the line extended from R [to the same point on AE where the line from F intersects it] because AE is perpendicular to the plane in which lines RE and
FE lie, since this plane is the plane of the [conic] section, and the two lines RE and FE are equal. Hence, both lines extended from R and F to a given point on line AE are equal.

6.30◉ It is therefore evident that the form of [any] point on AN will be reflected to R from a point [such as] that on line AE. And the same holds for any point lying on AN beyond N; if it is connected with F by a straight line, that line will intersect AE beyond E. It is also evident that the form of a point on AN will be reflected to R from a point on AE. From this, therefore, it is
evident that the form of line AN, as well as any [line] continuous with it, will be reflected to R from a straight line on the surface of cone ABG, and the same holds for every line extended from A at a slant to the cone’s axis.⁑

6.31◉ Let us draw ND [in figure 6.6.22a, p. 131, abstracted from figure 6.6.22].⁑ It will therefore intersect the periphery of the [conic] section because the
two points N and D lie in the plane of [that] section, and N lies outside the section[‘s periphery], whereas D lies inside the section[‘s periphery]. So let it intersect the periphery of the [conic] section at C, and since triangle AOH lies in the same plane, ND will lie in the same plane as triangle AOH.

6.32◉ [Point] C is therefore in the plane of triangle AOH, and the two points A and N lie in the plane of this triangle. Hence, points A, N, and C lie in the plane of triangle AOH. But points A, U, and C lie on the surface of the cone. Accordingly, points A, U, and C lie on the common section of the surface of the cone and plane AND. But this common section is a straight line. So points
A, U, and C lie in a straight line.

6.33◉ Extend AU directly to C, then, and extend RZ in a straight line. Accordingly, it will intersect OH [because O, Z, R and H all lie in the same plane of reflection]. Let it intersect at P. P therefore lies in the plane of triangle AOH. So extend AP, and let it continue in a straight line. It will therefore intersect ND at G, and since F lies below the plane tangent to the cone that
passes along line [of longitude] AZE, angle FED will be acute, whereas [adjacent] angle DEN is obtuse. Hence, [interior] angle ENC [of triangle NED] is acute [because it is smaller than opposite exterior angle FED, which is acute].

6.34◉ Furthermore, let line CZ’ be tangent to the [conic] section. It is clear, then, as [shown] in an earlier proposition, that angle DCZ’ is obtuse⁑ and that the perpendicular erected on CZ’ at C cuts angle DCZ’ and will intersect ED beyond D. So this perpendicular will intersect ED at S.

6.35◉ Hence, the perpendicular extended from N to the line tangent to the [conic] section [at the point where that perpendicular intersects the conic section] will intersect the [conic] section beyond C, that is, farther from E than C [lies from it], for these perpendiculars [i.e., NQ and ED] will intersect outside the periphery of the [conic] section [i.e., on the other side of that
periphery from N]. Hence, the perpendicular extended from N to the line tangent to the [conic] section will not cut angle DCZ’. It will therefore lie farther from NE than CD [does], and this perpendicular cuts ED beyond D.

6.36◉ So let the perpendicular dropped from N to the line tangent to the [conic] section be NQ. Also, RE intersects EN, and it intersects the periphery of the [conic] section and lies in its plane, while NQ [also] lies in the plane of the [conic] section. Hence, if RE is extended in a straight line, it will intersect NQ. Let it intersect at Y, then.

6.37◉ Plane AND intersects the plane of the [conic] section. Since point E lies outside plane AND because plane AND is not [in] the plane of the [conic] section [whereas E is], and because A lies outside the plane of the [conic] section, since AE is perpendicular to the plane of [that] section, whereas E lies on its periphery, then ND is the common section of plane AND and the plane of
the [conic] section, and NQ will intersect [that] section beyond C [i.e., on the opposite side from C and E]. Hence, NQ lies outside plane AND. Y therefore lies outside line APG.

6.38◉ Accordingly, if the center of sight lies at R, and if line AON lies on some visible object, then P will be the image of O, Y will be the image of N, and A will appear at its [actual] location, since it lies at the vertex of the cone.⁑ And the image of line AON will be the line passing through points A, P, and Y, but this line is convex because it lies outside [straight line] APG.

6.39◉ So let that [image-]line be APY, and it has already been shown that the forms of all the points on AN are reflected to R from AE. Therefore, the radial lines according to which those forms are reflected lie in the plane of triangle RZE, so all the images of [the points on] line AN lie in that plane.

6.40◉ Hence, convex line APY lies within that plane, and P lies closer to R than Y does, and the convexity of this image will be toward the center of sight, and it will [therefore] be of slight [apparent] convexity.⁑ Moreover, the [length along the] cross-section [AY] of this image will be slightly smaller than the line itself [i.e., AN of which it is the image]. Consequently,
the images of straight lines that are extended from the vertex of the cone at a slant to the axis are perceived by sight as convex in such a mirror, and the forms of these lines are reflected from straight lines among the lines extended along the cone’s longitude, and this is what we wanted to prove.

6.41◉ On the other hand, the forms of lines that are parallel to the width of a conical convex mirror are reflected from convex lines on the mirror’s surface, and the convexity of these lines is obvious, as [it is] in a convex cylindrical mirror, and for the same reason, and it will likewise be evident that the images of these lines will be quite convex and manifestly [so] to the [visual]
sense. Also, the center of sight will lie outside the plane that contains the convexity of the forms of these lines, and the cross-sections of the images of these lines will be considerably shorter than the lines themselves.

6.42◉ As to lines that are slanted between these two extremes, however, those whose orientation approaches that of lines extended along the length of the cone have slightly convex forms, whereas those that approach lines parallel to the width of the cone have forms that are clearly convex.

6.43◉ But also, curved lines that approach the vertex of the cone have smaller, narrower, and more convex forms, whereas those that approach the base of the cone have larger forms, according to what was demonstrated for convex spherical mirrors—i.e., that the smaller the mirror, the smaller the circles that fall on its surface—and so the images [falling on those smaller circles] will lie
closer to the center [of curvature], from which it follows that they will be smaller.

6.44◉ By the same token, sections that lie on a conical mirror toward the cone’s vertex are narrower and shorter [than those that lie farther from it], and so the image [within such a section] will be nearer the point where the normals dropped from the visible line to the lines tangent to the sections, which form the common section [of the plane of reflection and the plane tangent to the
mirror at that point], intersect, and so those images will be smaller.

6.45◉ On the other hand, the opposite holds for sections that lie toward the base of the mirror, so it happens that a form perceived in a convex conical mirror will take on a conical form, i.e., what lies toward the vertex of the mirror will be narrower, whereas what lies toward the base will be broader, and the convexity of a form along the width [of the mirror] will be evident.

6.47◉ Therefore, the misperceptions that occur in these sorts of mirrors are in every way like those that occur in convex cylindrical mirrors except for the conical shape of the form. And without exception the form of a visible object that is perceived by reflection will always take the shape of the surface of the mirror from which the form is reflected, and the reason for this is that
the image-location is invariably determined by the shape of the mirror’s surface and by the place where the normals intersect, so the [shape of the] mirror’s surface always plays a role in the shape [of the image] of the visible object that is perceived in the mirror. However, the compound misperceptions [arising] in this [sort of] mirror are identical to the [compound] misperceptions [occurring] in the previously discussed mirrors [i.e., convex spherical and
convex cylindrical].⁑

7.1◉ In these [mirrors], in fact, more [misperceptions] occur than in all the convex and plane mirrors,⁑ for what occurs in the latter occurs in these as well—i.e., a weakening of light and color and a variation in orientation and distance—for it is
reflection alone, not the shape of the mirror, that causes this [sort of variation]. [But] in addition, there is more variation in [image] size in these mirrors than in convex mirrors, for in convex [mirrors] an object will generally be perceived as smaller [than it actually is], whereas in concave [mirrors] it will sometimes be perceived as larger, sometimes as smaller, [and] sometimes as it actually is, and this happens according to how it changes position with
respect to the mirror as well as to the center of sight, as we will demonstrate in this chapter.

7.2◉ It also happens in these mirrors that a single visible object may appear as two, or three, or four, and this is not the case in plane and convex mirrors, for in those [kinds] a single visible object is perceived only singly, whereas in concave [mirrors, such is] not [the case].

7.3◉ Furthermore, the arrangement of the visible object’s parts is perceived in convex and plane mirrors as it actually is, whereas in spherical concave [mirrors it is perceived] otherwise in several situations,⁑ and this in two ways: namely, in convex spherical mirrors there is no deception in the perception that a single thing is single and the perception of the arrangement of its parts according to how it actually is, and since there is deception in regard to these aspects in spherical concave mirrors,
it is clear that nothing is perceived in these mirrors without deception, either invariably or at some time according to variation in the position [of the object vis-à-vis the mirror as well as the center of sight].

7.4◉ However, weakening of light and color as well as change in position and distance occur in these mirrors just as [they] invariably [occur] in the others, and they do so in every situation. But size, shape, and number are subject to deception in these mirrors in some situations, as we will demonstrate.

7.5◉ Concerning number it has been shown in chapter [2, book 5] on image [formation] that in concave spherical mirrors one object has one, two, three, or four images, and that the form of a visible object is always perceived at its [appropriate] image-location. However, one object perceived in concave spherical mirrors may be perceived as one, perhaps as two, perhaps as three, and perhaps
as four, which does not happen in convex and plane mirrors.

7.6◉ As to the arrangement of the visible object’s parts, it has also been claimed in chapter [2, book 5] on image [formation] that the form of a single [object-]point is reflected from the circumference of a [great] circle [on the mirror’s surface] and that visible objects whose images lie beyond or behind the center of sight, in front of it, or at the center of sight [itself] appear
blurred and not clear, and anything of this sort does not have the arrangement of parts that the visible object itself has. And here, as well, what obtains in these mirrors is other than what obtains in convex and plane mirrors. But the reasons for this phenomenon have been discussed in the chapter on image [formation].⁑

7.7◉ It thus remains [for us] to show that what is perceived in these mirrors may be perceived larger, smaller, or the same size [as the object itself], and that in certain situations it may be perceived inverted and in others erect, and that a straight object is perceived as concave, convex, or straight in mirrors of this sort, and that convex and concave objects are also perceived other
than they [actually] are [in this sort of mirror]. And these [misperceptions] also arise from a variation in the arrangement of the visible object’s parts, and we will demonstrate this in the following way.

7.8◉ [PROPOSITION 23] Accordingly, let there be a concave spherical mirror centered on A [in figure 6.7.23, p. 132], let it be bisected by a plane passing through its center, let it form [great] circle BG, let a line [AU] be extended within it at random, and let it be bisected at O.

7.11◉ So when the center of sight is at T, and when M and N lie on some visible object, the form of M will be extended along line MB and will be reflected along BT, and the form of N will be extended along NG and will be reflected along GT. The center of sight at T will therefore perceive points M and N [at locations] beyond points [of reflection] B and G, and [so it will perceive the
entire image of] line MN beyond arc BG.⁑

7.12◉ Also, since TE is perpendicular to AB, angle ABT will be acute. But angle MBA = angle ABT. Thus, TB > BM, so AT > BM, and they [i.e., lines AT and BM] are parallel. Consequently, [line of reflection] TB will intersect [cathetus] AM. Let them intersect at F, then. F is thus the image of M, and it will be demonstrated equivalently that [line of reflection] TG will intersect
[cathetus] AN. Let it intersect at Q, then. Q will thus be the image of N.

7.13◉ Let us then connect FQ, which is the cross-section of the image of MN, and since TE and TZ are equal, angles TA[E]B and TAZ[G] will be equal, lines TB and TG will be equal, lines BM and GN [will be] equal, and lines AM and AN [will be] equal. Moreover [given the similarity of triangles AFT and MFB], AF:FM = AT:BM, and AF:FM = AT:GN = AT:BM [because GN = BM], so AF:FM [= AT:GN] =
AQ:QN, and AM = AN. Hence, AF = AQ, so FQ is parallel to MN. Thus, FQ > MN. But FQ is the cross-section of the image of MN. Accordingly, if the center of sight is at T and MN lies on some visible object, the eye will perceive its form as larger than [object-line MN] is.

7.14◉ [PROPOSITION 24] Now [in figure 6.7.24, p. 132] let us duplicate circle BG, line AT[U], and lines AB, AG, and TB [as given in figure 6.7.23]. Let TK be perpendicular to the plane of circle BG at point T, and let us draw KA, KB, and KG. Thus, planes KBA and KGA intersect the sphere [of the mirror] at its center perpendicular to [the appropriate] planes tangent to its surface.⁑ Within these [planes], then, the form [of any given visible object] is reflected, and the two common sections between these two planes and [the surface of] the sphere form great circles from whose circumference the forms are reflected.

7.15◉ Let us then draw BM parallel to AK in plane BKA, and let it be shorter than AK. Let us draw AM, and let it be extended in a straight line, and extend KB until they intersect at F. Then draw NG in plane KGA, let it be parallel to AK, and assume it is equal to BM. Let us connect AN, let it be extended in a straight line, and extend KG in a straight line until they intersect at Q. Then
let us connect MN and FQ.

7.17◉ Hence, when the center of sight lies at point K, and when line MN lies on some visible object, the form of M will be extended along line MB and will be reflected along line BK in the plane of the circle passing through points B, A, and K, whereas the form of point N will be extended along line NG and will be reflected along line GK within the plane of the circle passing through
points G, A, and K.

7.18◉ And [so] point F will be the image of point M, while point Q will be the image of point N, and line FQ will be the cross-section of the image of [the entire line] MN. But we have already demonstrated [in paragraph 7.16] that line FQ > line MN, so when the center of sight is at point K, and when line MN lies on some visible object, the eye will apprehend the form of line MN on
line FQ. Therefore, it will perceive the form [of the visible object] as larger than the visible object [itself].

7.19◉ Accordingly, if we rotate the entire figure around line AU, while keeping [AU] itself stationary [to form the axis of rotation], point K will produce a circle that is perpendicular to line AU, and so every point beyond that point on that circle will be situated with respect to a line equivalent to line MN as K is situated with respect to MN.

7.20◉ Consequently, if the center of sight lies at any point on the circum-ference of this circle, and if a line equivalent to line MN lies on the surface of some visible object [that is similarly disposed], the eye will perceive the form of that line [as] larger [than the line itself]. Likewise, moreover, if we extend TK in a straight line and take some point on it other than K [as a
center of sight], and if we extrapolate at every stage from that point, which is equivalent to point K, the case will be like the case for point K.

7.21◉ On the basis of these two propositions [i.e., 23 and 24], therefore, it is evident that in concave spherical mirrors many objects are perceived [as] larger [than they actually are] in many situations.

7.22◉ [PROPOSITION 25] To continue, let AB [in figure 6.7.25, p.133] be a concave spherical mirror centered on E, and let us produce a plane passing through E, and let it form [great] circle AB [on the sphere]. Let us extend line EZ randomly from E to G, and from G let us drop GD perpendicular to the plane of circle AB, and let us mark point D on it at random. Then let us connect DE and
extend it to O, let us produce EB so that it forms an obtuse angle [DEB] with ED, and let us produce EA so that it forms an angle [AED] with ED equal to angle DEB. Let us then connect DA and DB. Accordingly, the planes of the two triangles DAE and DBE intersect one another along line DE, and the two acute angles DBE and DAE will be equal.

7.23◉ Now from B in the plane of triangle DBE let us produce a line [BO] forming an angle [EBO] with EB equal to angle DBE. Hence, this line intersects line DE, since angle BEO is acute, and the angle [EBO] at B is acute. So let it intersect at O.

7.24◉ From A let us also produce a line [AO] in the plane of triangle DAE that forms an angle [EAO] with AE equal to angle DAE. So let it intersect DE at O because the two angles AEO and BEO are equal [by construction], and because the angles at the two points A and B [i.e., EAO and EBO] are equal [by construction].

7.25◉ Let us then produce ET so that it forms a right angle with EB, and let us extend TE in the direction of E and BO in the direction of O, and let them intersect at H, and [so] TE = EH [insofar as triangles TEB and HEB are equal, by Euclid, I.26]. Let us likewise produce EK so that it forms a right angle with EA, let us extend it in the direction of E, and let us extend AO, and let
them intersect at L. Therefore, KE = EL.

7.26◉ Let us then connect TK and LH. They will thus be equal [because they lie within triangles KTE and HLE that are equal, insofar as KE = EL, TE = EH, and angle KET = vertical angle HEB]. Hence, if the center of sight lies at D, and if LH lies on some visible object, then D will perceive LH in mirror AB, and T will be the image of H [whose form is reflected from point B], K the image of
L [whose form is reflected from point A], and so TK will be the cross-section of the image of LH, and it is equal to it.

7.27◉ Consequently, if we rotate the entire figure, leaving HL stationary [as the axis of rotation], D will produce a circle, and if the center of sight lies at any given point on its circumference, it can perceive some visible object equivalent to line LH, and the image will be equal [in size] to it. And likewise, if the center of sight lies at O and TK is the visible object, the image
will be the same size as the visible object.

7.28◉ But yet, if the visible object is LH, if the eye is at D, and if TK is the image, the image will be inverted; [for] if H lies on the right [of object HL from D’s point of view its image] T will lie on the left [of image TK from that same point of view], whereas if H lies to the left, T will lie to the right, and if H lies above the line, T will lie below the line, and the same for
L.

7.29◉ Moreover, if the visible object is TK, if the center of sight is at O, and if the image is LH, the form will be erect, for image LH will lie beyond the center of sight, and it will be perceived ahead of the visible object, as was shown in chapter [2] on image [formation] in the fifth book, and the eye will perceive H, which is the image of T, along line BO, and L, which is the image
of K, along LO.⁑

7.31◉ [PROPOSITION 26] Now let us continue BH [in figure 6.7.26, p. 133] in a straight line and mark R on it[s extension], and let us connect RE. Angle REB will therefore be obtuse [since it is larger than HEB, which is a right angle, by construction].

7.34◉ Accordingly, if MR lies on some visible object, and if the center of sight is at D, NU will be the cross-section of the image of MR, and NU < MR. On the other hand, if the center of sight is at O, and if NU lies on some visible object, MR will be the image of NU, and it is longer than NU.

7.35◉ But if MR is the visible object, and if NU is its image [as viewed from D], then the image will be inverted, whereas if NU is the visible object and MR is its image [as viewed from O], the image will be correctly oriented, for if it lies beyond the center of sight, that image will appear ahead [of the object], and every point on the image will appear along a specific line among the
[corresponding] radial lines.⁑

7.36◉ [PROPOSITION 27] To continue, let us mark point Q on line OH [in figure 6.7.27, p. 134]. Let us connect QE, and let it continue to C. Let OF = OQ, and let us connect EF and let it continue to I. The two lines CE and EI will thus be longer than the two lines EF and QE, and [so] line CI > line FQ [in similar triangles EIC and QEF].

7.37◉ Hence, if the center of sight is at O, and if CI lies on some visible object, FQ will be the image of CI, and FQ < CI. Moreover, FQ will appear along the two lines AO and OB. Therefore, the form [of CI] will lie in front of the center of sight and will be smaller than the visible object [itself], and it will be properly oriented.

7.40◉ [PROPOSITION 28] Now let AB [in figure 6.7.28, p. 135] lie [on] a concave [spherical] mirror [with] G its center, let that mirror be bisected by a plane passing through its center, and let it form [segment] AB [of a great] circle. Let us extend line GD at random, let it pass to E on the side of G, let the center of sight be at E, and let T lie on the surface of the eye.

7.41◉ Then let us draw TH perpendicular to line ED, let ZT = TH, and let [the center of sight at] E perceive [the form of] H [by reflection] from A.⁑ Consequently, the two points A and H will lie on opposite sides of point G, for if they lay on the same side, the line extending from the mirror to A would not cut the angle that the two radial lines [of incidence and reflection] form.⁑

7.42◉ Let us then draw lines EA, AH, GA, and GH, and let GH extend in a straight line to K. Hence, the two angles at A [i.e., HAG and GAE] will be equal [by construction], and K [where cathetus HGK and line of reflection EA intersect] will be the image of H.

7.43◉ Let arc BD = arc DA, let us draw lines EB, BZ, and BG, and let us extend ZG to L. Thus, the two angles at B [i.e., ZBG and GBE] will be equal, and [the form of] Z will be perceived by the center of sight [according to reflection] from B, and L will be the image of Z.

7.44◉ Let us connect KL. KL will therefore be the cross-section of the image of ZH, and since [object-line] ZTH is perpendicular to DE [by con-struction], and since ZT = TH [also by construction], the two lines EA and AH will be equal [respectively] to the two lines EB and BZ, the two angles [HAG and GAE] at A are equal to the two angles [ZBG and GBE] at B, and line GH = line ZG.

7.46◉ Moreover, angle HGA is obtuse, and the two angles [HAG and GAE] at A are equal [by construction], so line GH > line GK, and likewise ZG > GL.⁑ Hence, line KL < [line] ZH [because of the similarity of isosceles triangles HGZ and GLK]. But KL is the cross-section
of the image of ZH. Therefore, line ZH will appear shorter than it actually is. Moreover, line ZH is [on] the viewer’s face [insofar as it is a cross-section of the eye that faces the mirror].

7.47◉ Therefore, if we rotate the circle at B [i.e., arc BDA] around ED, leaving EG[D] stationary [as the axis of rotation], it will produce a circle, and it will produce a circle on the mirror’s surface from the two points A and B. In addition, the position of center of sight E with respect to any line equivalent to ZH on that circle marked off by ZH and with respect to any arc
equivalent to arc AB on the segment [of the circle] that the two points A and B mark off on that circle will be equivalent to the position that center of sight E has with respect to line ZH and arc AB. And the proof will be the same whether we suppose the [object-]line to be longer or shorter than ZH.

7.48◉ From all of these conclusions, it is clear that the cross-section of the surface of the viewer’s face is perceived [to be] smaller than it [actually] is in the concave [spherical] mirror. So it follows that, if the center of sight lies at E, the viewer will perceive his face in such a mirror as smaller than it is, and since K is the image of H, while L is the image of Z, the image
will be inverted.

7.49◉ Accordingly, the center of sight at E will perceive the viewer’s form as such, i.e., it will perceive what lies to the right to the left, and [what lies] below above, and vice-versa. By the same token, if the center of sight lies at any point such that the center of [the mirror’s] curvature lies between it and the mirror’s surface, it will perceive its [viewer’s] form inverted, and
this is what we wanted [to demonstrate].⁑

7.50◉ It is therefore evident from these four propositions [i.e., 25-28] that in a concave [spherical] mirror [an object] is sometimes perceived as larger, sometimes smaller, and sometimes the same size [as the object itself], and [it] sometimes [appears] properly oriented, sometimes inverted.

7.51◉ Moreover, in chapter [2, book 5] on image [formation], we explained that in a concave [spherical] mirror the image will sometimes be single, sometimes double, sometimes triple, and sometime quadruple, and this same phenomenon occurs in the situations just discussed.

7.52◉ Hence, whatever yields an image that is larger than itself may yield others that are smaller or the same size, whereas whatever yields a smaller image may yield others larger or the same size, and whatever yields an image the same size [as itself may yield] a larger or smaller [one], and whatever appears upright [according to one image] may appear inverted according to another
image, and vice-versa. So it remains to analyze the forms of those things that are perceived in these sorts of mirrors.

7.53◉ [PROPOSITION 29] Accordingly, let AB [in figure 6.7.29, p. 136] be a [concave] spherical mirror, let us produce a plane bisecting that mirror through the center, and let it form [great] circle AB centered on E. In this circle let us draw two intersecting diameters, AEO and BED, and let the mirror not extend past arc BADO. Let us then select point Z at random on BE, let us select
point K on line AE, and let AK > KE. Then let us connect ZK, and let it continue to F. Let us also draw EF, and let angle EFG = angle EFZ.

7.54◉ Thus, since FK > KA, and since KA > KE [by construction], FK > KE.⁑ Angle FEK is therefore greater than angle EFK [by Euclid, I.19], so it is greater than angle EFG. Hence, line FG will intersect line KE.⁑ Let them intersect at G, then. Consequently, the two lines ZF and FG are
reflected at equal angles [ZFE and GFE], so K [where cathetus GEK and line of reflection ZKF intersect] is the image of G if the center of sight is at Z.

7.56◉ Let us take angle EHR = angle EHZ. Line HR therefore intersects line GF, and it [also] intersects line EG, so let it intersect line EG at R. Hence, the two lines ZH and HR are reflected at equal angles [ZHE and RHE], and L [where cathetus REL intersects line of reflection ZLH] will be the image of R [for center of sight Z]. I say, then, that the form of any point on line GR is
reflected to the center of sight Z from a point on arc FH, and from no other [arc].

7.57◉ The proof of this is [based on] both figures 27 and 28 in chapter [2] on image [formation] in book 5, where it has been shown that the two arcs AB and DO cannot be such that anything on line EO will be reflected from them to [center of sight] Z, and the mirror does not extend to arc BO.⁑ Consequently, only arc AD is left [for the reflection].

7.58◉ However, in the thirty-fifth proposition [of book 5] it has been shown that the form of any point on diameter EO is reflected at some point on arc AD, and in the thirty-sixth [proposition] of chapter [2, book 5] on image [formation] it was demonstrated that the form of point R is reflected to Z from only one point on arc AD.⁑ Therefore, the form of any point on line GR is reflected to Z from one point only on arc AD.

7.59◉ Let us take point C on line GR [in figure 6.7.29a, p. 136]. The form of C is therefore reflected to Z from one point on arc AD. I say, then, that that point will lie only on arc FH. For if such is not the case, let the form of C be reflected to Z from U, which lies on arc AF, and let us connect lines ZU, CU, GU, and EU.

7.60◉ Therefore, line GU > line GF, and ZU < ZF, so GU:ZU > GF:FZ. Hence, [GU:ZU] > GM:MZ [which = GF:FZ, by previous conclusions]. The line that bisects angle GUZ therefore intersects line ZM, so it [also] intersects ZE. Consequently, angle GUE < angle EUZ; so a fortiore angle CUE < angle EUZ, and the same holds for any [other] point on arc AU[F]. The form of C is
therefore reflected to Z from arc [D]HF only.

7.61◉ I say, furthermore, that it cannot be reflected [to Z] from arc HD. For if that were possible, let it be reflected from Q, which lies on arc HD [in figure 6.7.29b, p. 137], and let us connect lines ZQ, CQ, RQ, ZR, and EQ, and let us extend EH to N. Therefore, line ZQ > [line] ZH, and line QR < [line] HR, so ZQ:QR > ZH:HR, which = ZN:NR [by Euclid, VI.3, since angles ZHE and
RHE were constructed equal]. The line that bisects angle ZQR therefore intersects line NR, so it intersects line ER. Consequently angle RQE > angle EQZ, so a fortiore angle EQC > angle EQZ. The same result follows for any [other] point on arc HD, so the form of C is not reflected to Z from arc HD or from arc AF.

7.62◉ However, it has already been shown that it absolutely must be reflected from arc AD. Consequently, the form of C is only reflected to Z from some point on arc FH. Accordingly, let it be reflected from T [in figure 6.7.29c, p. 137], and let us connect lines CT, ET, and ZT. Therefore, since T lies between the two points F and H, line ZT will lie between the two lines Z[K]F and Z[L]H.
Line ZT therefore intersects line KL. Let it intersect it at I, then. I is therefore the image of C, and C has no image other than I.

7.63◉ And it will be demonstrated in this way that the image of any point on line GR will be a point on line KL. Thus, KL is the image of [the entire line] GR, and KL is a straight line because it is a segment of the circle’s diameter. GR is also a straight line because it too is a segment of the circle’s diameter. Thus, in [concave] spherical mirror AB, Z perceives the form of GR
according to its proper [left-to-right] orientation, and this is what we wanted [to prove].

7.64◉ [PROPOSITION 30] Now let us copy the [previous] figure, and let us circumscribe two random arcs on both sides of line GR, namely, GNR and GQR [in figure 6.7.30, p. 138], and let arc GNR not intersect line GH. Let us select point M at random on line GR. The form of M is therefore reflected to Z from [some] point on arc FH. Let it be reflected accordingly from T, and let us connect
lines ZT and MT.

7.65◉ Hence, the two angles ZTE and ETM are equal, so line MT will intersect arc GNR. Let it intersect that arc at N, then, and let us extend line TM on the side of M. It will therefore intersect arc GQR, so let it intersect at point Q. Next let us connect NE and extend it in a straight line. It will therefore intersect ZT below line KL. So let it intersect that [line] at I. Then let us
connect QE and extend it in a straight line. Accordingly, it will intersect ZT above KL. Let it intersect that line at C, then.

7.66◉ Consequently, since the two angles [ZITE and NTE] at T are equal, I will be the image of N, and the two points K and L are the images of the two points G and R. The image of arc GNR is therefore a line passing through points K, I, and L, i.e., line KIL. But line KIL is convex with respect to the eye, and arc GNR is convex with respect to the mirror. Z will therefore perceive the
form of convex line GNR [as] a convex line.

7.67◉ Moreover, since the two angles [ZCTE and QTE] at T are equal, C will also be the image of Q, and line LCK, which is concave with respect to the eye, will be the image of arc GQR, which is concave with respect to the mirror’s surface. Z will therefore perceive the form of concave arc GQR [as] a concave line.

7.69◉ [PROPOSITION 31] Now let there be a concave [spherical] mirror containing great circle ABD [figure 6.7.31, p. 138], let G be the center, let us draw line BG at random, and let us cut from it line GT longer than its half. Let us then draw line ETZ from T orthogonal [to BG], and let both ET and TZ be equal to TG. Let us connect ET, EG, and GZ.

7.70◉ Let us then circumscribe a circle around triangle EGZ. It will therefore intersect circle AB at two points, for point T is the center of this [new] circle, and TG > TB [by construction]. So let this circle intersect circle AB at the two points A and D, and let us connect lines GA, GD, EA, EB, ED, ZA, ZB, and ZD.

7.71◉ Accordingly, since the two lines ET and TZ are equal [because they are both equal to TG, by construction], the two lines EB and BZ will be reflected at equal angles [i.e., EBG and GBZ subtended by equal arcs]. Also, since the two arcs EG and GZ are equal, the two lines EA and AZ are reflected at equal angles [EAG and GAZ subtended by those equal arcs], and the two lines ED and DZ
will [also] be reflected at equal angles [EDG and GDZ subtended by equal arcs].

7.72◉ Since GT > TB [by construction], GE > EB. Hence, angle EBG > angle EGB, and angle EGB is half a right angle. The two angles EGB and EBG thus sum up to more than a right angle. Consequently, angle BEG < a right angle, whereas angle EGZ is a right angle. The two lines EB and GZ will therefore intersect outside the circle [EGZ] on the side of BZ. So let them intersect at
M.

7.73◉ In addition, since ED lies within angle MEG, it will intersect line GM. Let them intersect at L. And since GB passes through the center of circle EGZ, segment AEG < a semicircle. Therefore, angle AEG is obtuse, whereas angle EGZ is right. The two lines AE and ZG will thus intersect on the side of EG. Let them intersect at F, then. Accordingly, if the center of sight is at E, and
if Z lies on some visible object, points M, L, and F will be images of Z. Z is thus perceived at three places.

7.74◉ To continue, let us draw a line at random from E to arc DZ, and let it be EK [see inset to figure 6.7.31]. Let us connect GK, let it intersect arc DZ at K, and let us connect lines KZ and GK. Therefore, since arcs EG and GZ are equal, the two angles EKG and GKZ will be equal. [Let us extend GK to point K’ on the mirror, and let us connect EK’ and ZK’]. Consequently, angle EK’G >
angle GK’Z.⁑ Let angle GK’N = angle EK’G, then. Consequently, the two lines EK’ and K’N will be reflected at equal angles. Let us then extend EK’ to Q. Q will therefore be the image of N with respect to [center of sight] E.

7.75◉ Let us now imagine a plane passing along line MGF perpendicular to circle ABD [as represented in figure 6.7.31a, p. 139], let us draw a line from Z in this plane perpendicular to GZ, and let it extend on both sides [of Z]. Accordingly, let it be CZR. Let us then take G as a centerpoint, and let us produce arc CNR of a circle with radius GN. It will therefore intersect line CR at two
points, and let them be C and R. Let us connect lines GC and GR. They will thus lie in a plane perpendicular to plane ABG [of the mirror]. Let us then extend GC and GR in a straight line, and at point G let us produce the arc of a circle with radius GQ. It will thus intersect the two lines GC and GR. Let it intersect [them] at S and O.

7.76◉ Hence, since the plane of circle ABD [on the mirror] is perpendicular to the plane of the two lines GC and GR, the two angles EGS and EGO will be right. Both planes EGS and EGO will thus be perpendicular to plane SGO, and both of those planes cut a great circle on the mirror that is equivalent to circle ABD. From the counterpart of point K’ in the [great] circle formed by plane EGC,
then, two lines between points E and C are reflected at equal angles.⁑

7.77◉ Moreover, lines GC and GR are equal, lines GS, GQ, and GO are equal, and Q is the image of N, S is the image of C, and O is the image of R. Therefore, the image of arc CNR, which is convex with respect to the mirror, is arc SQO, which is concave with respect to the center of sight.

7.78◉ Meantime, L is the image of Z, and the two points S and O are the images of C and R. Consequently, the image of straight line CZR is a line passing through points S, L, and O, and such a line is concave with respect to the center of sight.

7.79◉ Let us now draw the line passing through points S, L, and O, and let us extend line EG to H. Accordingly, if the mirror does not reach the two points B and H, but one of its two limits lies between the two points B and D, while the other lies inside of H [i.e., between H and D], and if the center of sight lies at E, while the two lines RZC and RNC lie on some visible object, the
form of straight line RZC will be concave, i.e., SLO, and the form of convex arc RNC will also be a concave line, i.e., SQO. Furthermore, straight line RCZ will have a single image, and arc RNC will [also] have a single image.

7.80◉ Now let us extend BG to I, and let us connect lines EI and IZ. Those two lines will therefore be reflected at equal angles, and EI will intersect FG, so let it intersect at T’. Hence, T’ will be the image of Z. Points M, L, T’, and F will therefore [all] be images of Z. And if the mirror extends beyond the two points A and I, while the center of sight lies at E, and if the viewer
faces the mirror on the side of arc AI, he will perceive the entire arc IDA.

7.81◉ Consequently, Z will appear at four places, i.e., at L, M, T’, and F, and he will see the two points R and C at the two points S and O, so line RZC will have four concave images. One passes through points S, M, and O, i.e., line SMO; a second will pass through points S, L, and O, i.e., line SLO; a third will pass through points S, T’, and O, i.e., line ST’O; and a fourth will pass
through points S, F, and O, i.e., line SFO.

7.82◉ From this proposition it is therefore clear that in concave [spherical] mirrors a straight line is perceived as concave, a convex [one] is also perceived as concave, and a straight [line] has several concave forms.⁑

7.83◉ [PROPOSITION 32] Now let there be a concave [spherical] mirror through whose center a plane passes, let it produce [great] circle ABG [in figure 6.7.32, p. 140], and let D be its center. From D let us draw a line at random, let it be DG, and let it extend beyond the circle. From point D let us extend a line in the plane of the circle perpendicular to line DG, and let it be DA. Let
us then cut from right angle ADG a small sub-angle GDE at random such that the difference between right angle [ADG] and angle ADE consists of several [increments of] angle EDG,⁑ and let us
bisect angle ADE with line DB. Let us also cut [from angle ADG] a sub-angle [ADZ] equal to angle EDG, and let us extend a line from D that forms a right angle with DB [i.e., BDT], and let it be DT.

7.84◉ Now let us extend AD on the side of D, let it form DK, and let us extend a line [ZH] from Z that forms with ZD an angle [DZH] equal to angle KDT. This line will therefore intersect DA because the two angles KDT [which = DZH, by construction] and ADZ sum up to less than two right angles. So let them intersect at H. Angle ZHD is thus equal to angle ZDT.⁑

7.85◉ Then from Z let us extend line ZL to form an angle [HZL] with ZH equal to obtuse angle BDK. Accordingly, the two angles LZD and BDZ sum up to less than two right angles, so line ZL will intersect [line] DB.⁑ Let it therefore intersect at L.

7.86◉ Let us then connect LH, and let us form circle DHL around triangle HLD. It will therefore pass through Z because the two angles LZH and LDH sum up to two right [angles, by Euclid, III.22]. Consequently angles LHZ and LDZ are equal because they are subtended by the same arc [LZ]. But angle ZHD = angle ZDT [by previous conclusions], so it follows that angle LHD = angle LDT. But angle
LDT is right [by construction], so angle LHD is right.

7.88◉ Let us now draw lines HB, HF, FM, FZ, and FB. Angle BHD will thus be acute, while angle GDH will be right [by construction]. Therefore, line HB will intersect line DG outside the circle. Let them intersect at Q, then. HF will therefore also intersect DG outside the circle, so let them intersect at N.

7.89◉ Let us then extend FB until it intersects arc LZ. Accordingly, let it intersect at R, and let us connect RM. Angle FRM, which lies on the circumference [of circle ZDF], is thus subtended by arc FM, and angle FBM > angle FRM, but angle FBM lies on the circumference of [circle] ABG. Therefore, if line BM is extended on the side of M, it will cut a larger arc on circle ABG than the
counterpart FM [it cuts on circle ZDF].

7.91◉ Hence, if BM is extended in a straight line on the side of M, it will cut an arc on circle ABG beyond point F that is greater than arc FG. Line BM will therefore intersect line DG between the two points G and D. Accordingly, let it intersect at O. Let us then extend line FM, and let it intersect DO at U; let us also extend BM on the side of B, and let it intersect arc LR at C. Let
us then connect CD.

7.92◉ Accordingly, because angle BFZ lies on the circumference of [circle] ABG, angle BFZ will be half of angle BDZ [by Euclid, VI.33]. But angle BDZ is several times larger than angle ZDA [by construction], so angle R[B]FZ is several times larger than angle ZDH[A]. Consequently, arc RZ is several times the size of arc ZH, and arc CZ > arc RZ, so arc CZ is several times the size of arc
ZH.

7.95◉ In addition, angle FMD exceeds angle MUD by only [one increment of] angle EDG [by Euclid, I.32]. Therefore, angle MUD > angle MOG, so angle MOU [adjacent to MOG] > angle MUO [adjacent to MUD]. Hence, line MU > line MO [since MU subtends the larger angle]. And since arc ZHD = arc FMD [by previous conclusions], the two angles HFD and MFD will be equal. The two lines HF and FU
will therefore reflect at equal [angles], and likewise HB and BO will reflect at equal [angles]. Consequently, Q is the image of O, and N is the image of U [from the perspective of H, as the center of sight].

7.96◉ From M let us then extend a line parallel to line HQ, let it be MS, and let us also extend a line from M parallel to line HN, and let it be MP. Thus, since angle HND > angle HQD, angle MPO > angle MSO. P therefore lies between the two points S and U, and because angle HDN is right [by construction], angle HND will be acute. Accordingly, angle MPD is acute, so angle MPS
[adjacent to it] is obtuse. Line MS is therefore longer than [line] MP.

7.98◉ Now that these points have been established, let us redraw the circle and finish the proof so as to avoid adding lines and confusing letter-designations. Accordingly, let ABG [in figure 6.7.32b, p. 142] be the circle in the second version [of figure 6.7.32, p. 140], let D be its center, and let us draw line DQ. Let DU [in the second version] be equivalent to DU in the original
version, and let DO [in the second version] be equivalent to DO in the original version. Also, let DQ [in the second version] be equivalent to its counterpart in the original version, and likewise for DN.

7.99◉ Let us then draw DH’ perpendicular to DQ [as well as] to the plane of the circle, and let DH’ be equivalent to its counterpart [DH] in the original version. Angle H’DQ will therefore be right, and the [great] circle that [the plane containing] H’DQ produces in the mirror will be among the circles [like ABG] within which a form is reflected. Furthermore, the arc that lines H’D and DQ
measure off [in the great circle produced on the mirror by plane H’DQ] will be equal to arc AG in the original circle. And from the two points on this [arc] that are equivalent to the two points B and F [on arc AB in the original circle] lines from the two points U and O will be reflected at equal angles [to point H’]. Q will therefore be the image of O [for center of sight H’], and N [will be] the image of U.⁑

7.100◉ From U let us produce a line perpendicular to line DU within the plane of circle ABG, and let it be ZUE. Let D be the center, and [from it] let us produce the arc of a circle with a radius of DO. It will therefore intersect line ZUE at two points. Accordingly, let it intersect at Z and E, and let it form arc ZOE. Let us then draw DZ and DE, and extend them beyond the circle. At a
radius of DQ, let us produce arc TQK around D. It will therefore intersect the [extension of the] two lines DZ and DE at T and K. Let us then draw TK. Accordingly, it will intersect line DQ at L.

7.101◉ Consequently, since H’D is perpendicular to the plane of the circle, both angles H’DT and H’DK will be right. Moreover, both planes H’DT and H’DK produce a [great] circle on the mirror’s surface, and the arc on it that lies between the two lines H’D and DT will be equal to the arc lying between HD and DQ [in the original figure—i.e., 6.7.32], and the same holds for the arc between
the two lines H’D and DK. In addition, both lines DZ and DE are equal to line DO. These two arcs [cut on the mirror’s surface by planes H’DZ and H’DE] are therefore of the kind from which lines will be reflected at equal angles to the two points Z and E.⁑ Furthermore, the two lines DT and DK are equal to line DQ, so point T is the image of Z, and [point] K is the image of E.

7.102◉ Since, moreover, lines DT, DQ, and DK are equal, and since lines DZ, DO, and DE are equal, DT:DZ = QD:DO = KD:DE. But, as we showed in the previous theorem [i.e., in paragraph 7.97 keyed to figure 6.7.32], QD:DO > ND:DU. Therefore, DT:DZ > ND:DU, and the same holds for KD:DE.

7.103◉ In addition, since the two lines ZD and DE are equal, and since the two lines DT and DK are equal, line TK will be parallel to line ZE. Therefore, DT:DZ and KD:DE will [both] be as LD:DU. Hence, LD:DU > ND:DU, so line LD > line ND. N therefore lies between L and U. But N is the image of U, and the two points T and K are the images of Z and E. As a result, the image of
straight line ZUE is the line that passes through points T, N, and K. But the line that passes through these points is convex, from which it is clear that in concave [spherical] mirrors a straight line sometimes appears convex in certain situations.

7.104◉ Now let us take some point M at random on line ZU [in figure 6.7.32d, p. 143], and around M as center let us produce arc RUF with radius MU. This arc will therefore intersect arc ZOE⁑ in two points. Accordingly, let it intersect at R and F, let us draw lines DR and DF, and let them pass in a straight line until they intersect arc TQK at C and I. The plane containing the two lines H’D and DC will therefore produce a [great]
circle on the mirror from whose circumference lines from R will be reflected at equal angles, and by the same token the plane containing the two lines H’D and DI will produce a [great] circle on the mirror from whose circumference lines will be reflected to F [at equal angles]. C is therefore the image of R, I is the image of F, and N is the image of U.⁑

7.105◉ Consequently, the image of arc RUF is the line passing through C, N, and I, but this line will be convex, whereas arc RUF is concave with respect to the mirror’s surface. Therefore, when the center of sight is at H’, and when any of the lines ZUE, ZOE, or RUF lies on some visible object, straight line ZUE will be perceived [as] convex, convex line ZOE will be perceived [as]
concave, and concave [line RUF is perceived as] convex. Consequently, if each of the lines ZUE, ZOE, and RUF has [only] one image, the form of those lines will be just as we showed. But if they have additional images, they may be similar to the other images [i.e., the ones just discussed], or they may be different.

7.107◉ So the forms of visible surfaces are perceived [as] other than they [actually] are in these sorts of mirrors, for straight lines exist only in flat surfaces, and when a straight line that exists in a plane surface is perceived [as] convex or concave, the surface in which it lies will be perceived [as] convex or concave. Accordingly, when the eye perceives convex, concave, and
straight lines other than they [actually] are, it will perceive the surfaces in which they lie other than they are.

7.108◉ From the foregoing, then, it is clear that in everything that is perceived in concave [spherical] mirrors, a misperception occurs, but in certain cases it occurs in every situation, without exception, whereas in certain [cases] it occurs in a specific situation. Moreover, compound misperceptions arise in these mirrors just as in the case of compound illusions [in the other
mirrors],⁑ and this [is what] what we wanted to demonstrate.

8.1◉ In these [sorts of mirrors] the same things happen as happen in concave spherical [mirrors], for the misperceptions that arise from reflection [by itself] occur [in these mirrors], i.e., the weakening of light and color and the variation in situation and distance that occur in all mirrors. Moreover, variation in size occurs in these mirrors in the same way as it happens in concave
spherical mirrors. Also, one visible object appears [as] one, or [as] two, or [as] three, or [as] four, and [it appears] properly oriented or reversed in various circumstances, and a flat object appears concave or convex. So let us show how the size and number of a visible object may vary in these mirrors, as well as how it appears properly oriented or reversed in the way that we demonstrated [these phenomena] in spherical concave mirrors.

8.2◉ [PROPOSITION 33] Let us recapitulate the first of the two diagrams provided in [the analysis of] misperceptions [occurring in] convex cylindrical mirrors [i.e., figure 6.5.18, p. 127, redrawn as figure 6.8.33, p. 144], and [let us use] the same letters. Now in that proposition [i.e., proposition 18, in conjunction with proposition 17, pp. 190-193 above] it was shown: that lines EG and
GT, EB and QB, and EA and AH are reflected at equal angles; that lines EO, HA, BQ, and TG intersect at O;⁑ that line ABG is a straight line extended along the longitude of the mirror; that lines GZ, BL, and AD are perpendicular to the plane tangent to the [mirror’s] surface and passing along line ABG; that line ABG is perpendicular to the plane containing triangle EBO; that line TQ = [line] QH, and [line] AB = [line] BG; that S, C, and I are the images of H, Q, and T; that C lies nearer point E than [straight] line SI; that [straight] line SI lies in the
plane of triangle UHT; that the two lines UH and UT are equal; that the two lines US and UI are equal; and that the two lines ES and EI are equal.

8.3◉ Let us draw CU, and let it intersect SI at F. It will therefore bisect this line [i.e., SI] because HT is bisected at Q [by construction], and CU will lie in the plane of triangle CUE, which lies in the plane of the circle [passing through] B parallel to the base of the mirror.⁑ Q will therefore lie in the plane of triangle CUE, and C lies in triangle CEI. Hence, C lies on the common section of these two planes. But this [common] section is line EB; so C lies on a straight line [with] EB.⁑

8.4◉ Moreover, the two lines HU and TU [in figure 6.8.33] lie outside the two points D and Z [on the axis], for the two lines HU and TU are the normals passing from H and T to the two lines that are tangent to two sections [on the cylinder’s surface] on whose periphery points A and G lie [i.e., the two elliptical sections formed on the mirror’s surfaces by planes of reflection TGE and
HAE]. Consequently, the plane of triangle UHT lies outside axis DLZ.

8.5◉ Even if the axis is extended to infinity, however, no point on it will lie in the plane of triangle UHT, for if it did, then if it were to be connected in a straight line with some point on line HT, the plane in which that straight line and line HT lay would be the plane of triangle UHT, and that plane would be the one in which the two parallel lines HT and DZ lie. Hence, the plane
containing the two lines HT and DZ [supposedly] forms the plane of triangle HUT, so the axis will lie in the plane of triangle HUT.

8.6◉ But the axis is parallel to line HT by construction, and the axis [supposedly] intersects the two lines HU and TU. Moreover, line TH lies in [i.e., passes through] the plane of triangle UEH, which is the plane of reflection [for object-point H and reflection-point A], and the common section of this plane and the surface of the mirror is some [elliptical] section. Therefore, plane EUH
intersects the axis of the cylinder in one point, that is, D, as we showed before [in proposition 18]. And if the axis intersects line HU, the point of intersection with line HU will lie in the plane of triangle UEH. But there is no point in this plane other than D through which the axis might pass. Therefore, line HU intersects the axis at D. But it has already been shown that HU intersects it at a point beyond D, which is impossible.⁑

8.7◉ Consequently, axis DZ lies outside the plane of UHT and nearer to point E than plane UHT [i.e., between E and plane UHT]. The plane in which lines HT and DZ lie is therefore nearer to point E than plane UHT. Moreover, C lies in the same plane as HT and DZ because it lies on line QL, and QL lies in the same plane as HT and DZ. Therefore, C lies nearer to point E than [do] S and I. But
C lies in a straight line with EB. If, therefore, EB is extended toward B, it will reach C, so let it reach C.

8.8◉ Now that these points are established, I say that if line SI, which is parallel to the mirror’s axis, lies on some visible object, if the center of sight lies at O on the concave side of the cylinder, and if the reflecting surface is a concave surface, then SI will be perceived by O in concave mirror ABG, and its images will vary according to how its distance from the axis varies.

8.9◉ The proof of this [claim] lies in the fact that angle EBM is acute, so [vertical] angle LBC is acute. Moreover, line EBC lies in the plane of the circle [passing through] B, and LB is [on] the diameter of this circle. Hence, EB intersects the circle, so CB lies inside the mirror’s concavity.

8.10◉ By the same token, OB lies inside the mirror’s concavity because angle OBL is acute, and the two angles OBL and CBL are equal, since they are equal to the two angles EBM and QBM, while LB is perpendicular to the plane that passes through B tangent to the cylinder. The form of C thus passes along CB and reaches B, and it is reflected along BO and perceived by the center of sight at
O.

8.11◉ Furthermore, when we discussed convex cylindrical mirrors in chapter 5 [of book 4, paragraph 5.18, in Smith, Alhacen on the Principles, 332], we showed that the plane tangent to the cylinder at G will lie below E. Therefore, EG intersects the tangent plane, so it intersects the line tangent to G on the periphery of the [elliptical] section [formed on the mirror by the plane of
reflection]. As a result, it intersects the [elliptical] section and falls inside it; so it will fall inside the concavity of the mirror. The two lines OG and GI thus lie inside the concavity of the mirror, whereas ZG is perpendicular to the plane tangent to the cylinder at G, and the two angles OGZ and IGZ are equal. Hence, the form of I passes along IG, reaches G, is reflected along GO, and is perceived at O along line GO. So too, [the form of] S passes along SA
and is reflected along AO.

8.12◉ But when we dealt with misperceptions [arising] from convex cylindrical mirrors, we demonstrated [in proposition 16, lemma 5] that the two lines HU and TU are normal to the two planes tangent to the [elliptical] sections passing through the two points A and G. Therefore, the image of S lies on line HU. Moreover, OA is the radial line extending from the center of sight to the point
of reflection, so the image of S lies on OA. H [where the two lines intersect] is therefore the image of S, and it is shown in this way that T is the image of I.

8.13◉ Let us then connect CL. Accordingly, since [the form of] C is reflected to O from the periphery [of the circle passing through] B, [its] image Q will lie on line CL. And OB is the radial line extending between the center of sight and the point of reflection, so the image of C lies on line OB. Consequently, the image of C lies at the intersection of [cathetus] QL and [line of
reflection] OB [i.e., at Q].

8.14◉ When we dealt with images in concave spherical mirrors in chap-ter [2, book 5] on image [formation], however, it was shown [in proposition 32, in Smith, Alhacen on the Principles, 446-448] that the image of a point whose form is reflected from the concavity of a [great] circle [on the mirror’s surface] may intersect the radial line linking the center of sight and the point of
reflection beyond the circle, or between the center of sight and the circle, or at the center of sight [itself], or behind the center of sight, or CL may be parallel to OB.

8.15◉ In that [same] chapter [on image formation], moreover, it was shown that the image might consist of a single point, or of two, or of three, or of four.⁑ So the image of C might lie [at some point between B and Q] on BQ, or perhaps beyond Q, or perhaps on BO, or perhaps at O, or perhaps behind O.⁑ Moreover, the
image of C might consist of a single point, or of two, or of three, or [of] four.

8.16◉ Accordingly, if the image of C lies at Q, then HT will be the cross-section of SI’s image. So, if all the images of [points on] SI lie on line HT, its form will be a straight line. If not, however, it will be nearly straight because its midpoint lies on a straight line between two endpoints.⁑ Nevertheless, if the image of C lies beyond Q, the image of SI will be somewhat concave with respect to the center of sight. And if the image of the visible [point C] lies on line BO [i.e., in front of Q], then the image of SI will be convex with respect to the center of sight.

8.17◉ Moreover, if the image of C consists of several points, the image of C will lie on several lines, all of whose endpoints converge at the two points H and T, and their midpoints are distinct and separate. In addition, HT forms the cross-section of image SI, no matter how that image is formed, and [this] cross-section is common to all of its images if it has several images, and line
HT [on the image] is longer than [line] SI [on the object] by some amount.⁑

8.18◉ It is therefore evident that, when straight lines parallel to the axis of a concave cylindrical mirror lie on some visible object, the image of any [one of them] may be straight or concave, and it may consist of a single [line] or [of] several.

8.19◉ [PROPOSITION 34] Now let us recapitulate the second diagram [provided in the analysis] of misperceptions in convex cylindrical mirrors [i.e., figure 6.5.19, p. 128, which accompanies proposition 19]. In this proposition [represented by figure 6.8.34, p. 147, abstracted from figure 6.5.19], it has been shown: that the two lines EB and BH are reflected at equal angles; that the two
lines EG and GT are reflected at equal angles; that HB and TG converge at L; and [that] HB forms an acute angle with BO. Consequently, HB intersects the plane tangent to the surface of the cylinder at B, so BL lies inside the concavity of the cylinder. And the same holds for GL, as well as for the two lines BR and GY.

8.20◉ Moreover, the two angles LBD and DBR are equal, and the two angles LGD and DGY are equal. Hence, if RY lies on some visible object, if the center of sight lies at L, and if the concave surface of the cylinder is polished [and therefore reflective], the form of R is extended along RB and reaches B. It will then be reflected along BL and will reach L, and it will be perceived by L.
Moreover, line HU [i.e., the cathetus dropped from R through the mirror’s surface] is perpendicular to a line tangent to the [elliptical] section from whose periphery the two lines RB and BL will be reflected. Therefore, H is the image of R. Likewise, it will be proven that the form of Y is extended along YG and is reflected along GL, and its image is T.

8.21◉ Let us now draw KU [in figure 6.8.34a, p. 148].⁑ Accordingly, it will intersect RY at M. M therefore lies in the plane passing through the axis and through L, for L and K lie in that plane, so KU lies in that plane. Moreover, since the two points M and L lie in a plane passing through the cylinder’s axis, the form of M will be reflected to L within that plane. Line AZ is the common section of the cylinder’s surface and the plane passing through its axis and through L, so the form of M will be
reflected to L from [a point on] AZ.

8.22◉ Let us then connect EM, which lies in this plane. EL also lies in this plane, and E lies above the plane tangent to the surface of the cylinder along line AZ. Hence, if AZ is extended in a straight line on the side of Z, it will intersect the two lines EM and EL. Accordingly, let it intersect EM at I and EL at N. N therefore lies between the two points E and L because L lies inside
the concavity of the cylinder, whereas N lies on the cylinder’s surface, and E lies above the [surface of the] cylinder.

8.23◉ Furthermore, in the proof based on this diagram, it was shown that circle BZG lies halfway between [the plane passing through] line HT [parallel to the base of the cylinder] and the plane passing through E parallel to the base of the cylinder. In addition, the perpendicular [EX’] that passes from E through AZ lies in the plane passing through E parallel to the cylinder’s base.
Therefore, the perpendicular passing through E to line AZN falls outside triangle EIN and on the side of N. Consequently, angle EIN is acute; so [vertical] angle MIA is [also] acute.⁑

8.24◉ So let us extend MQ [in figure 6.8.34b, p. 149] from M perpendicular to AI. Q will therefore lie beyond I with respect to N [i.e., below I on AZA’]. And let us extend MQ on the side of Q, and let us cut off QS equal to QM. S will therefore lie beyond the surface of the mirror and outside its concavity, while L will lie inside its concavity.

8.25◉ Let us then draw LS. Accordingly, it will intersect NQ at F, and from F let us extend FX parallel to QM. It is therefore perpendicular to AN and [lies] in the plane passing through the axis and through L, so it is a diameter of the circle [produced on the mirror’s surface by the plane] passing through F parallel to the cylinder’s base. Therefore, line XF is perpendicular to the
plane tangent to the cylinder and passing along AZ.

8.26◉ Let us connect MF. Accordingly, it will be equal to FS, and [so] the two angles [FMQ and FSQ] at M and S will be equal [because triangles FMQ and FSQ are equal, by Euclid, I.4]. Moreover, because XF is parallel to MS, the two angles [LFX and MFX] at F will be equal to the two angles [FSQ and FMQ] that are at S and M.⁑ The two lines MF and FL are therefore reflected at equal angles, and XF is perpendicular to the plane tangent to the mirror’s surface at F. So the form of M is extended along MF and is reflected along FL, and its image will be S.

8.27◉ Moreover, since the two lines RY and HT are parallel and [therefore] perpendicular to the plane passing through the axis and through L (because HT was posited as such [in proposition 19]), the two planes passing through the two lines HT and RY [perpendicular to the axis] will be parallel. Since, moreover, RY is perpendicular to the plane passing through the axis and through L, the
plane [consisting] of the two lines RM and MS will be perpendicular to the plane passing through the axis and through L. Furthermore, MS will be the common section of these two planes [i.e., RMS and ELDS in figure 6.8.34b], and since AQ lies in the plane [ELDS] passing through the axis and is perpendicular to MS, which is the common section of the plane [ELDS] passing through the axis and the plane [consisting] of the two lines RM and MS, AN will be perpendicular
to the plane [consisting] of the two lines RM and MS.

8.28◉ But line AN is parallel to the axis of the cylinder, so the axis of the cylinder is perpendicular to the plane containing the two lines RM and MS. Therefore, this plane is perpendicular to the axis of the cylinder. S therefore lies in the plane passing through line RY perpendicular to the axis of the cylinder.

8.29◉ But line HT lies in a plane perpendicular to the axis of the cylinder and parallel to the plane passing through line RY. Hence, S lies outside [and above] HT and nearer to L than HT. In addition, the two points H and T are the images of R and Y, and point S is the image of M, so the image of line RMY is the line passing through H, S, and T.

8.30◉ Such a line is curved, however, because S lies outside HT, and a curved line HST must pass through points H, S, and T. And since HT lay beyond the convex [side of the] cylinder, according to construction, HT will lie beyond the surface of the mirror with respect to L. In addition, we have already shown that S lies beyond the concavity of the mirror with respect to L, so the entire
line HST lies beyond the concavity of the mirror’s surface. Moreover, L lies inside the concavity of the mirror, so L lies outside the plane containing line HST. Therefore, the curvature of line HST will appear clearly to the eye at L.

8.31◉ Furthermore, since F lies on the surface of the cylinder, while TH lies beyond the cylinder, and since TH lies in the plane of triangle LHT, line LFS will be higher than the plane of triangle LHT. Line LS will therefore be higher than the two lines LH and HT with respect to the center of sight at L. Consequently, S is higher than the two points H and T, so line HST will appear
concave to the center of sight at L.

8.32◉ [PROPOSITION 35] To continue, let us cut the cylinder with a plane slanted to its axis, but do not let it pass through the entire axis [so as to cut the cylinder along a line of longitude]. Accordingly, it will form an [elliptical] section. Let it therefore be ABG [in figure 6.8.35, p. 150].⁑ But in the first of the propositions concerning concave cylindrical [mirrors] it was demonstrated that in the plane of any [elliptical] section on a cylinder there will be a normal to the plane
tangent to the cylinder from whose endpoints forms are reflected.⁑ So let that normal be GZ[A], let BE[K] be perpendicular to the line tangent to the periphery of the [elliptical] section at B, and let B lie near G. BK will therefore intersect normal GZ, and it will form an acute angle with it. Accordingly, let it intersect at E. Angle BEG will therefore be acute [by proposition 16, lemma 5].

8.33◉ From G let us extend line GD parallel to line BK. Hence, angle DGE will be acute, so GD will lie inside the concavity of the cylinder. Let us then take angle EGL equal to angle EGD. GL will thus intersect BE at L. And let us mark M on line LE [inside the elliptical section]. Angle MAG will therefore be acute because AM lies inside the [elliptical] section.

8.34◉ Let us then take angle GAD equal to angle GAM. Therefore, AD will intersect GD, since the two angles [GAD and AGD] that are at A and G are acute. So let them intersect at D. AD will therefore intersect BK. Let it then intersect at T.

8.35◉ Consequently, if BK lies on some visible object, and if the center of sight lies at D, the form of L will be seen at G because the form of L will be reflected to D from G, and DG is parallel to normal LB [by construction].⁑ Meantime, the form of M is seen at T because the form of M is reflected to G from A, and T is the image of M.

8.36◉ Now let a plane pass through D parallel to the base of the cylinder [to form the circle represented at the bottom of figure 6.8.35]. Accordingly, it will intersect the plane of ABG and will form circle COR on the surface of the cylinder. The plane of this circle will therefore intersect BK, for it intersects GD, which is parallel to it [by construction]. Let it therefore intersect
BK at K, and let point H be the center of circle CR. Let us then draw DH, and let it pass to R. Let us also draw KH, and let it pass to C.

8.37◉ Hence, the form of K is reflected to D from the periphery [of the circled centered on H] within arc RC, as was shown in [the analysis of] images [formed] in [great] circles [within concave spherical mirrors].⁑ So let it be reflected from O, and let us draw KO, DO, and HO. The angles [DOH and KOH] at O are therefore equal, and
[reflected ray] DO will intersect [cathetus K]HC at N. So N is the image of K.

8.38◉ Let us then connect KD. Accordingly, KD will be the common section of circle RC and [elliptical] section ABG, since the two points K and D lie in both planes, for there is nothing except line KD in plane ABG of the [elliptical] section that is [also] in the plane of circle RC. G therefore lies outside the circle, and likewise T, and both lie in the plane of the [elliptical]
section.

8.39◉ N, meanwhile, lies in the plane of the circle, and the form of LMK passes through points G, T, and N, and [so] the line that passes through these points is curved. However, the plane of the [elliptical] section is slanted with respect to the cylinder’s surface, so the [major] axis of the [elliptical] section does not pass along the entire axis of the cylinder, nor is it parallel to
the cylinder’s base.

8.40◉ From this and the previous two propositions it is therefore evident that straight lines parallel to the axis of the cylinder, as well as those parallel to its base, and also those that are slanted with respect to its surface may appear curved, or straight, or reversed. Furthermore, since T is the image of M and N the image of K, the form of MK will be reversed.

8.41◉ In addition, if the line also lies in the plane of a circle parallel to the cylinder’s base, and if the plane of that circle passes through the center of sight, the image may be the same size [as its object] and properly oriented, or it may be reversed, as was claimed in [propositions 25-27 of] the seventh chapter of this book [dealing] with images in [great] circles [on concave
spherical mirrors].

8.43◉ [PROPOSITION 36] Now let us recapitulate the diagram for the third proposition [dealing] with misperceptions in concave spherical mirrors, leaving the letters as they are [in figure 6.8.36, p. 151, which combines figures 6.7.26 and 6.7.27]. Let BZA be a circle on the surface of a concave cylindrical mirror, and let the center of sight be at D [on DG, which is perpendicular to the
plane of BZA]. It will therefore lie outside the circle’s plane, and the two lines EA and EB will [each] be perpendicular to a plane tangent to the cylinder’s surface [at points A and B]. In addition, the plane of triangle DGE will be perpendicular to the plane of the circle because DG is perpendicular to the plane of the circle.

8.44◉ Hence, the plane of triangle DGE passes through the entire axis as well as through D, whereas neither plane DBO nor plane DAO, which intersect along line DO, passes through the entire axis. Moreover, there is nothing but E on the cylinder’s axis in either plane, E being the circle’s center. And each of the planes DBO and DAO forms an [elliptical] section on the cylinder’s surface,
and forms are reflected from these [elliptical] sections at the two points A and B.

8.45◉ The form of R is therefore reflected to D from B, whereas the form of M is reflected to D from A, and NU will be the cross-section of the image of MR, and it is shorter than MR. Likewise, the [forms of the] two points H and L are reflected to D from the two points A and B, and TK will be the cross-section of LH’s image, and it is the same size as TK. Finally, CI will be the
cross-section of FQ’s image, and it is longer than FQ. All of these images, moreover, will be reversed.

8.46◉ But if the center of sight lies at O, and if lines CI, TK, and NU are the visible objects, the opposite will obtain, for in that case the cross-section of the image [FQ] of CI will be shorter than CI, whereas the cross-section of the image [MR] of NU will be longer than NU, and the cross-section TK [of LH’s image] will be the same size as it, and these images will all be properly
oriented. All these points were shown in the preceding chapter.

8.47◉ Furthermore, when either endpoint of any of these [lines] has a single image, and when any intermediate point [on that line] has several images, that line will have as many images as the intermediate point has. If, moreover, one endpoint or the other [of the line] has several images, and if the intermediate point has one, then the line will have as many images as the endpoint has.
And if one endpoint or the other has several images, and if the intermediate point has several images, the line will yield images according to the greatest number [as pointed out in note 130, p. 256]. This will be shown as was shown for images in concave spherical mirrors.

8.48◉ Hence, in concave cylindrical mirrors misperception occurs in all respects as it occurs in concave spherical mirrors, that is, concerning the shapes of visible forms, concerning the sizes and number of their images, and concerning their proper orientation or reversal, along with the misperceptions that apply to reflection [itself]. And the misperceptions in these cases will be as
they are in the aforementioned mirrors, and these are the points we wanted to demonstrate in this chapter.

9.1◉ In these [sorts of mirrors] the misperceptions that occur are those that occur in concave cylindrical mirrors. Indeed, the weakening of color and light, as well as variation in location and distance, occur in these mirrors as in all [other kinds of] mirrors, for the cause of this is reflection [itself]. In addition, a multitude of images arises in these mirrors, just as in concave
cylindrical and spherical mirrors, as was claimed in chapter [2, book 5] on image [formation]. What happens in these mirrors is also like what happens in concave cylindrical [mirrors], i.e., what is straight appears convex, or it appears concave.

9.2◉ The demonstration of this is that straight lines that extend along the length of the mirror and pass through the cone’s vertex, as well as those that are near these [lines in orientation], appear convex, or they appear concave, or perhaps [they appear] straight.

9.3◉ [PROPOSITION 37] The demonstration of this point is like the demonstration [given] for concave cylindrical mirrors [in proposition 33, pp. 221-224 above], for if we recapitulate the second diagram concerning misperceptions in convex conical mirrors [i.e., the top diagram of figure 6.6.22a, p. 131, from which figure 6.9.37, p. 152, has been abstracted], we will find the cross-section
of the image of straight line [AN] placed toward that mirror, which, in that case, is [curved line] A[P]Y inside the concavity of the conical mirror, and we will find the point that is below the plane tangent to the cone and passing along the line [AZE] from which the form of the straight line [AN] is reflected to the center of sight, which is F in that case.

9.4◉ If [that] point is the center of sight, all the points that are on the cross-section of the image will be reflected to point F, and the images of the two endpoints A and Y [on object-line APY] will be the endpoints of straight line AN, and the image-location of intermediate [point P] on AY will vary. And this will be demonstrated by the same train [of logic] we followed in the proof
[provided] in the first proposition [dealing] with concave cylindrical mirrors [i.e., proposition 33].⁑

9.5◉ From this it is clear that, if AY lies on some visible object, and if F is the center of sight, the image may appear convex, or it may appear concave.⁑ And it is also
evident from the second proposition concerning misperceptions in concave cylindrical mirrors [i.e., proposition 34] that lines placed along the width of a [concave conical] mirror will appear concave with a remarkable curvature and that images of straight lines that lie in planes passing through the axis and the center of sight will be straight.

9.6◉ [PROPOSITION 38] Now with the same letters, let us recapitulate the third figure concerning misperceptions in concave spherical mirrors [as given in figure 6.8.36, p. 151, which combines figures 6.7.26 and 6.7.27]. If, therefore, some point [i.e., E] lies on the axis of the cone, and if the two lines EA and EB [passing through that point] are perpendicular to planes tangent to the
cone (and this is possible because they are equal, since they can form two equal acute angles with the axis), then when [each of] these two lines is perpendicular [to a plane tangent to the mirror], and when the center of sight is at D, the plane containing lines GE and ED will pass through the entire axis as well as through the center of sight.

9.7◉ Furthermore, both planes [containing] DAM and DBR will be inclined to the axis of the cone, and [so] the common sections of those two [planes and the mirror’s surface] will be conic sections. The form[s] of points R, H, and Q will be reflected to D from B, and the forms of points L, M, and F are reflected to D from A. Hence, if lines MR, LH, and FQ lie on some visible surface, and if
the eye is at D, then NU will be the image of MR, TK will be the image of LH, and CI will be the image of FQ.

9.8◉ Thus, the image [NU] of MR will be shorter than [MR] itself, the image [CI] of FQ will be longer than [FQ] itself, and the image [TK] of LH will be the same size as [LH] itself, and all the images will be reversed.

9.9◉ If, moreover, the center of sight is at O, and NU, TK, and CI are on the surfaces of visible objects, their images will be MR, LH, and FQ. Accordingly, the image [FQ] of CI will be shorter than [CI] itself, the image [MR] of NU [will be] longer [than NU itself], and the image [LH] of TK will be the same size [as TK itself].

9.10◉ And these images will be properly oriented, for these images will lie behind the center of sight and will be perceived facing the center of sight along [direct] radial lines.⁑ Consequently, points M, L, and F are perceived along line [of reflection] AO, whereas points R, H, and Q are perceived along [line of reflection] OB, and so their form will be reflected with proper orientation.

9.11◉ From what we have claimed in this chapter, therefore, it is clear that straight lines sometimes appear convex in these mirrors, sometimes concave, and sometimes straight, and [they] sometimes [appear] longer, [sometimes] shorter, and [sometimes] the same size [as they actually are], and [they sometimes appear] properly oriented, and [sometimes] reversed.

9.12◉ In chapter [2, book 5] on image [formation], moreover, we showed that in mirrors of this kind every visible point sometimes has one image, sometimes two, or three, or four. Therefore, misperception occurs in everything that is perceived in this sort of mirror, just as in concave cylindrical [mirrors], and compound misperceptions also occur in these as in the rest of the mirrors.
Examples and proof of these [kinds of compound misperceptions] are as [they can be found] in plane mirrors. And we intended to explain this in this chapter. Now, however, let us end the sixth book.